11institutetext: Alibaba Groups, Hangzhou 311121, China
11email:
<EMAIL_ADDRESS>
# An Emotion-controlled Dialog Response Generation Model with Dynamic
Vocabulary

Shuangyong Song  Kexin Wang  Chao Wang  Haiqing Chen  Huan Chen

###### Abstract

In response generation task, proper sentimental expressions can obviously
improve the human-like level of the responses. However, for real application
in online systems, high QPS (queries per second, an indicator of the flow
capacity of on-line systems) is required, and a dynamic vocabulary mechanism
has been proved available in improving speed of generative models. In this
paper, we proposed an emotion-controlled dialog response generation model
based on the dynamic vocabulary mechanism, and the experimental results show
the benefit of this model.

###### Keywords:

Response Generation Emotion Analysis Generative Model.

##  1 Introduction

Chatbots aim to provide users multidimensional human-like services such as
shopping guide, chitchat and entertainments. Those kinds of services can
significantly help improve users’ experience and satisfaction. To create a
chatbot capable of communicating with a user at the human level, it is
necessary to equip the machine with the ability of perceiving and expressing
emotions.

For expressing appropriate emotions in chatbot responses, we linguistically
build emotional mappings between user questions and chatbot responses, and
then generate emotional responses with an emotion-controlled text generation
model. Compared to retrieval-based chatbots, generation-based chatbots can
mostly achieve greater coverage, and generate more proper responses that could
have never appeared in the corpus.

However, utilizing response generation model in real online systems has two
typical risks: the first one is that the content of generated response is not
entirely relevant to the user question, and the second one is that the running
speed of response generation models should be further improved to meet the
demand of real online systems. In [1], a dynamic vocabulary seq2seq (DVS2S)
model has been proposed and it can well solve above two risks, since it can
especially eliminate abundant noise words from the generation vocabulary,
which benefit both the generation speed and the relevance between responses
and questions.

In this paper, we try to realize a dynamic vocabulary based emotion-controlled
response generation model, which aims to generate emotional responses with
high quality and high speed for our chatbot, an industrial intelligent
assistant designed for creating an innovative online shopping experience in
E-commerce.

##  2 Our Model

Figure 1: Model architecture of DV-ERG.

1) Emotion Mapping between Questions and Responses: We first utilize LEAM
(Label-Embedding Attentive Model) [2] model to realize an emotion
classification on user questions, and then we linguistically build emotional
mappings between questions and responses, such as a mapping from an ‘abusing’
question to an ‘aggrieved’ or ‘regretful’ response.

2) Seq2Seq Model Training: For real application in online systems, with high
QPS requirement, we just employ typical Bi-GRU (Gated Recurrent Unit) as
encoder and GRU [7] as decoder, with an attention mechanism, instead of very
complex models. Employment of typical simple response generation models also
shows in other online systems, such as XiaoIce [3]. The objective function is
given below, and compared to a normal function, ours is meanwhile supervised
by emotional state e (in formula 1). In our training dataset, all the
$\left\langle Question,Response\right\rangle$ pairs comply with the above
mapping rules, and the e is set according to the emotion types of responses.
In the prediction, e is randomly chosen from candidate mapping emotions of
question emotion.

3) Vocabulary Model Training: Words are separated into function words and
content words, and the key is to predict the probability distribution P of
content words being selected in the decoder step. We take $\beta_{I(c)}$ as
converter from $\left\langle h,e\right\rangle$ to P, where the h is the hidden
state of the encoder, and the training task is to optimize $\beta_{I(c)}$.

4)Joint Fine-tune: We jointly fine-tune the Seq2Seq model and the vocabulary
model to further optimize the emotional response generation loss.

##  3 Experiments

1) Dataset collection & Implementation: We collect 132,118 frequently asked
emotional user questions from the online log of our commercial chatbot, and
manually labeled 1 to 3 corresponding emotional responses to each question.
Finally we got 308,618 QA pairs as training dataset of the seq2seq model. In
experiments, we use the 300-dimension pre-trained word embeddings, and we set
hidden size to 128 both in encoder and decoder, and learning rate to 0.001.

2) Baselines: We considered the following baselines: 1) S2SA: a standard
seq2seq model with an attention mechanism [4]; 2) TA-S2S: the topic-aware
seq2seq model proposed in [5]; 3) CVAE: recent work for response generation
with a conditional variational auto-encoder [6]; 4) DVS2S: the dynamic
vocabulary seq2seq model which allows each input to possess their own
vocabulary in decoding [1]. With the fine-tune step, we compare 3 different
ways: 1) no fine-tune (NO-ft); 2) just fine-tune the vocabulary model training
step (ft-target); 3) fine-tune both Seq2Seq and Vocabulary (ft-both).

Models | BLEU-2 | Recall | VocSize | Greedy | Average | Extreme | Distinct1 | Distinct2 | si-QPS
---|---|---|---|---|---|---|---|---|---
Baselines | S2SA | 2.79 | 94.21 | 50K | 38.80 | 41.88 | 22.10 | 0.512 | 0.415 |  68
TAS2S | 3.05 | 89.22 | 50K | 38.85 | 43.30 | 34.59 | 0.588 | 0.433 |  54
CVAE | 3.02 | 85.55 | 50K | 40.05 | 46.22 | 30.38 | 0.587 | 0.428 |  35
DVS2S | 3.21 | 93.01 | 4K | 42.11 | 49.38 | 37.77 | 0.595 | 0.646 |  94
DV-ERG | NO-ft | 3.38 | 92.16 | 4K | 42.01 | 48.10 | 38.61 | 0.614 | 0.670 |  92
ft-target | 3.83 | 92.16 | 4K | 41.79 | 48.15 | 38.50 | 0.605 | 0.655 |  92
ft-both | 3.71 | 90.20 | 6.8K | 41.31 | 48.09 | 38.70 | 0.604 | 0.658 |  92

Table 1: Evaluation results.

Input | Target emotion | Models |  Model reply | Labeled reply
---|---|---|---|---
Then wait patiently! | satisfied | Base |  My patience | Yeah, well, thank you. You’ve warmed me up!
NO-ft |  My patience
ft-target |  Mm-hmm, my little cute
ft-both |  Ok, mine.
Bullshit! | aggrieved | Base |  I don’t know | I’ve done my best.
NO-ft |  I don’t know
ft-target |  Sorry, I was wrong.
ft-both |  Oh oh.
Sing a song | non-emotional | Base |  Five-star red flag fluttering in the wind, what a resound-ing victory song! | No need to know
NO-ft |  I sing
ft-target |  OK
ft-both |  Sing what?

Table 2: Example results of different varieties of proposed DV-ERG.

3) Metrics: In this paper, we refer the evaluation metrics used in [1], and
further add several metrics: BLEU: Since the questions and answers in our
chatbot are almost short, we just consider BLEU-2. Recall: Only the R@1 metric
is relevant in the case of binary classification, which we use here. VocSize:
the size of the vocabulary. si-QPS: maximum QPS that a single model instance
can support. DISTINCT-1 & DISTINCT-2 Following [1, 5], we employ the Distinct1
and Distinct2 to measure how diverse and informative the generated responses
are. Besides, 3 embedding-based metrics [8] are used: Greedy, Average, and
Extreme.

4) Experimental results: Table 1 gives the evaluation results on different
metrics, and we can see ft-target gets the best performance on both BLEU and
si-QPS (si-QPS as 92 means about 10.87ms per query). For a real online
chatbot, those two metrics are more important than other metrics, so we choose
ft-target as the final online response generation model in our chatbot -
AliMe. We qualitatively analyze DV-ERG with some examples from the test data.
Table 2 shows several emotional generation results with our models, we can see
that most of the generative results are shorter than manually labeled results,
this is a common problem of generative models, since short results are ‘safer’
than long sentences in the model training step. However, with emotional
empathy and more focused vocabularies, generative results are sometimes even
better than manually labeled results. For example, with a user question as
‘Then wait patiently!’, the manually labeled response is ‘Yeah, well, thank
you. You’ve warmed me up!’. This response is with no problem, but it is better
when user get the generated response ‘Mm-hmm, my little cute ’ with ft-target
DV-ERG. Another example: with a user question as ‘Sing a song’, the manually
labeled response is ‘No need to know’, and this is just a so-so response. This
time, the ft-target DV-ERG generate a response as ‘Sing what?’, and this is a
more reasonable one.

##  4 Conclusion

In this paper, we proposed an emotion-controlled response generation model
based on the dynamic vocabulary mechanism, which can be practically applied to
online chat-bots, considering its experimental efficiency and effectiveness.
In the future, we will investigate how to apply the emotion analysis
technologies and dynamic vocabulary technique to more modules in online
chatbots.

## References

  * [1] Wu Y, Wu W, Yang D, Xu C, Li Z. Neural Response Generation With Dynamic Vocabularies. In AAAI 2018, pp. 5594-5601.
  * [2] Wang G, Li C, Wang W, et al. Joint Embedding of Words and Labels for Text Classification. In ACL 2018, pp. 2321–2331.
  * [3] Shum H-Y, He X, Li D. From Eliza to XiaoIce: challenges and opportunities with social chatbots. Frontiers of IT & EE 2019(1): 10-26.
  * [4] Vinyals, O., and Le, Q. 2015. A neural conversational model. arXiv preprint arXiv:1506.05869.
  * [5] Xing, C.; Wu, W.; Wu, Y.; Liu, J.; Huang, Y.; Zhou, M.; and Ma, W.-Y. Topic aware neural response generation. In AAAI 2016, pp. 3351-3357.
  * [6] Zhao, T.; Zhao, R.; and Eskenazi, M. Learning Discourse-level Diversity for Neural Dialog Models using Conditional Variational Autoencoders. ACL 2017, pp. 654-664.
  * [7] Cho, K., Merrienboer, B. V., Gulcehre, C., Bahdanau, D., Bougares, F., Schwenk, H., Bengio, Y. Learning phrase representations using RNN encoder–decoder for statistical ma-chine translation. In EMNLP 2014, pp. 1724– 1734.
  * [8] Liu, C-W., Lowe, R., Serban, I., Noseworthy, M., Charlin, L., Pineau, J. How NOT To Evaluate Your Dialogue System: An Empirical Study of Unsupervised Evaluation Metrics for Dialogue Response Generation. In EMNLP 2016, pp. 2122-2132.


# A Nanographene Disk Rotating a single Molecule Gear on a Cu(111) Surface

H.-H. Lin1, A. Croy1, R. Gutierrez1, C. Joachim2 and G. Cuniberti1 1 Institute
for Materials Science and Max Bergmann Center of Biomaterials, TU Dresden,
01069 Dresden, Germany
2 GNS and MANA Satellite, CEMES-CNRS, 29 rue J. Marvig, 31055 Toulouse Cedex,
France
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We perform molecular dynamics simulations to study the collective rotation of
a graphene nanodisk functionalized on its circumference by tert-butylphenyl
chemical groups in interaction with a molecule-gear hexa-tert-
butylphenylbenzene supported by a Cu(111) surface. The rotational motion can
be categorized under-driving, driving and overdriving regimes calculating the
locking coefficient of this machinery as a function of external torque
applied. Moreover, the rotational friction with the surface of both the
phononic and electronic contributions is investigated. It shows that for small
size graphene nanodisks the phononic friction is the main contribution,
whereas the electronic one dominates for the larger disks putting constrains
on the experimental way of achieving the transfer of rotation from a graphene
nanodisk to single molecule-gear.

,

  * August 2021

##  1 Introduction

Progress in the miniaturization of solid states gears down to the nanoscale[1,
2, 3] is calling for the study of gear trains and of the transfer of rotation
from a given gear diameter to the next (up and down in scale) [4]. We are now
reaching the experimental stage where a single molecule-gear 1 nm in
diameter[5] will have to be rotated by the smallest possible solid state
nanogear with diameter in the 10 nm range[2, 3]. Already the transfer of
rotation from one molecule-gear to the next has been observed along a
molecule-gear train where each rotation axle along the train is a single
metallic ad-atom[6]. The nanotechnology handicraft required to construct a
nanoscale machinery to transfer rotation from a solid state nanogear to a
molecule is under active exploration[7]. It encompasses gear thickness
compatibility, atomic scale surface preparation and preservation after
transferring the solid state nanogear on it, rotation axle stability for both
the solid state and the molecule gears and the means to activate the rotation
of the solid state nanogear to observe the rotation of the molecule-gear.
There is here also the fundamental question of the interactions of both the
solid state nanogear and the molecule-gear of this machinery with the
supporting surface and how this impacts the torque required to rotate both
gears starting from the solid state one.

Figure 1: A schematic illustration of the (a) topview (b) sideview for a tert-
butylbiphenyl functionalized graphene disk with radius and hexa-tert-
butylbiphenylbenzene on Cu(111) surface.

In this work, we present a theoretical study of the transmission of rotation
between a single layer graphene nanogear and a single molecule-gear both
adsorbed on the same atomically defined surface (Fig. 1). The molecule-gear is
an hexa-tert -butylphenylbenzene (HB- BPB) molecule with 6 tert-butylphenyl
teeth[8]. The 10 nm in diameter graphene nanodisk is supposed to be also
equipped with tert-butylphenyl teeth. With this design, the chemical structure
of the teeth on both gears is identical. The thickness of the graphene
nanodisk (hereafter called the master) is exactly compatible with the
molecule-gear chemical structure and with its physisorption height on the
Cu(111) surface. This compatibility is usually difficult to reach
experimentally since electron beam nanolithography processes ultimately lead
to a 5 to 10 nm gear thickness[2, 3]. A newcomer in the clean room, the
focused He+ beam nanolithography (HIM) will normally lead to the possibility
of sculpturing one after the other graphene monolayer nanodisks in high
vacuum[6].

With a circumference nanodisk edge having essentially sp2 carbon like dangling
bonds at the end of the HIM process, this is opening the way to chemically
bond the tert-butylphenyl teeth also on the graphene nanodisk. As presented in
Fig. 1 and for a 10 nm diameter graphene nanodisk master, we have determined
the optimal number of tert-butylphenyl around its circumference to ensure a
nice transmission of rotation. We found that 20 tert-butylpheny chemical
groups are working well. Other master diameters will certainly require to
adapt again its edge chemical structure. There is also a central 1.5 nm hole
in the master Fig. 1 to figure out the location of the rotation axle which for
example can be a mold of copper ad-atoms. The center-of-mass separation
between this master and the HB-BPB molecule-gear is 6.5 nm. This was also
optimized for optimum transmission of rotation. We have already observed
experimentally the importance of the pm precision of this center of mass
separation relative to the chemical structure of the teeth[8, 6]. To model the
Cu(111) supporting surface, we consider hereafter a three layers Cu(111)
surface (with size 15.79 nm $\times$ 12.766 nm $\times$ 0.23 nm) made of 10080
atoms with periodic boundary conditions in the lateral $x$ and $y$ directions.
We first preposition the master and its HB-BPB molecule-gear 0.5 nm above the
Cu(111) surface. To avoid a net translation of this surface after the
application of a torque to the master (action-reaction principle), we fix the
bottom Cu layer and allow only to relax the upper two layers. Finally, and
before starting the calculations, a geometry optimization of the complete Fig.
1 structure was performed using the conjugate gradient method to start with an
optimized surface conformation for the Fig. 1 machinery.

We first demonstrated how with the Fig. 1 design, a 5 nm graphene nanodisk can
effectively transfer a regular rotation movement to the interacting HB-BPB
molecule-gear. This is functioning in a restricted range of torque applied to
the master required to fight against the surface friction of the master but
not to destabilize the molecule-gear chemical structure itself by using a too
large torque. In a second step, we present a more detail theoretical study of
the effect of mechanical surface friction as a function of the master diameter
and compare it with its electronic friction on the Cu(111) surface. In
conclusion, we discuss how the rather simple nanoscale mechanical machinery of
Fig. 1 can be operated experimentally.

##  2  Model system and computational approach

In this section, we introduce the details concerning our molecular dynamic
(MD) simulations, describe the near rigid-body approximation used and recall
the concept of locking coefficient well known at the macroscopic scale to
study gearing effects along a train of gears. To carry out the MD simulations
on the nanoscale machinery of Fig. 1, we exploit the Large-scale
Atomic/Molecular Massively Parallel Simulator (LAMMPS)[9]. For the force
fields, we choose the Reactive force field (ReaxFF)[9] parametrized for Cu, C
and H. To fix the rotational axles, we connect a stiff spring with spring
constant $k=1600$ N/m (1000 eV/Å2) to each gear center-of-mass. In this way we
avoid the atomic construction of the axles especially of the Cu ad-atoms
cluster mold, for the master axle of rotation is not essential at this stage.
To determine the temperature of the supporting surface, the upper two Cu
layers are subject to the canonical ensemble implemented here by a Nosé–Hoover
thermostat[10, 11] reaching a surface temperature $T$ = 10K (the actual single
molecule mechanics experiments with a low temperature scanning tunneling
microscope (STM) are performed at about 5 K[5, 6]). To extract from the MD
simulation the different conformation angles of the deformable master nanodisk
and of the molecule-gear, the nearly rigid-body approximation[12, 13] was
exploited. For both the master and the molecule-gear in Fig. 1 , we define the
reference atomic scale structures by a set of coordinates
$\\{{\bm{r}^{0}_{k_{\alpha}}}\\}$, where $k_{\alpha}$ runs through all atoms
of the gear $k_{\alpha}$ with $\alpha=1,2$ for the master and the molecule-
gear, respectively. For example, we can choose the reference atomic structure
to be the one at time $t=t_{0}$ with its optimized coordinates for the master
and molecule-gear before performing the MD simulation
$\\{{\bm{r}_{k_{\alpha}}(t_{0})}\\}$. Supposing that the coordinates at time
$t$ is given by {${\bm{r}_{\alpha k}(t)}$}, the deformation in all frames are
here assumed to be sufficiently small such that the structure
{${\bm{r}_{\alpha k}(t)}$} can still be mapped to the reference atomic
structure $\\{{\bm{r}_{k_{\alpha}}(t_{0})}\\}$ via a rigid-body rotational
transformation matrix $\bm{R}^{\alpha}(\theta_{\alpha}(t),\bm{n}_{\alpha}(t))$
defining the rotational axes $\bm{n}_{\alpha}(t)$ and the angles
$\theta_{\alpha}(t)$. The rigid-body transformation matrix
$\bm{R}^{\alpha}(\theta_{\alpha}(t),\bm{n}_{\alpha}(t))$ can be found by
minimizing the deformation using the quaternions method[14, 13]. To
characterize the mechanical transmission of rotation between the master and
the molecule-gear, we use the concept of locking coefficient[13], well known
for macroscopic gears, and defined here by:

$L_{\alpha}=\frac{\langle\omega_{\alpha}\rangle}{\omega_{R\alpha}}\;,$ (1)

where $\langle\omega_{\alpha}\rangle$ denotes the average angular velocity of
gear $\alpha$ with $\omega_{R\alpha}$ the terminal angular velocity of the
rigid-body corresponding to this gear $\alpha$. This quantity provides a
measure of the ability to transfer a rotation between the two gears in Fig. 1.
Note that, for perfectly interlocked rigid gears, the coefficients
$L_{\alpha}$ must be equal to unity. For soft gears or a not perfect
transmission, we expect $L_{\alpha}$ to be much smaller than unity as
discussed below.

Figure 2: The angle displacement multiplied with number of teeth $N_{1}=20$
for graphene disk ($N_{2}=6$ for HB-BPB) within 100 ps with external torque
(a) $\tau_{ext}=12.8$ nN$\cdot$Å and (b) $\tau_{ext}=48$ nN$\cdot$Å. The blue
(red) line represents the. (c) The locking coefficient $L_{1,2}$ with respect
to external torque ranged from 0 to 160 nN$\cdot$Å, where region I, II and III
represent underdriving, driving and overdriving phases, respectively.

##  3  Results and discussions

###  3.1 Rotational transmission

To investigate the transmission of rotation in the machinery shown in Fig. 1,
we apply an external torque $\tau_{ext}$ on the master and follow how the
corresponding angular moment is transferred (or not) from this master to the
HB-BPB molecule. The $N_{\alpha}\theta_{\alpha}(t)$ variations in time for a
small $\tau_{ext}=12.8$ nN$\cdot$Å and an intermediate $\tau_{ext}=48$
nN$\cdot$Å torques are presented in Fig. 2 (a) and (b) with $N_{1}=20$ and
$N_{2}=6$ respectively. For a small torque (Fig. 2 (a)), we found only small
amplitude oscillations of both gears around their equilibrium conformation.
The applied torque is not large enough to fight against the master surface
friction on Cu(111) and to permit the molecule-gear to pass over the energy
barrier of the potential energy surface on Cu(111) at a $T$= 10 K surface
temperature. Notice also the damping effect of the Cu(111) surface is more
effective to reduce the oscillation amplitudes of the master as compared to
the ones of the molecule-gear (see also section III.B below). On the other
hand, for the selected intermediate torque in Fig. 2(b), we found a perfectly
interlocked rotation respecting the classical condition
$N_{1}\theta_{1}+N_{2}\theta_{2}=0$[15]. Let us recall here that to reach such
a result, we have equipped the edge of the master with exactly 20 equi-spaced
tert-butylpheny chemical groups. To understand this perfect transmission of
rotation, we have also plotted the locking coefficient as a function of torque
in Fig. 2(c). To calculate Eq. 1 for the master, we have calculated its
terminal velocity $\omega_{R1}$ using:

$\omega_{R1}=\frac{\tau_{ext}}{\gamma_{1}},$ (2)

where $\gamma_{1}=2.83\times 10^{-29}$ kg$\cdot$m${}^{2}\cdot$s-1 (extracted
from $\tau_{ext}/\langle\omega_{1}\rangle$ in the same MD simulation) is the
damping coefficient corresponding to rigid disk on the Cu surface. For
$\omega_{R2}$, we can take the time derivative of the interlocked condition
$N_{1}\theta_{1}+N_{2}\theta_{2}=0$, leading to

$\omega_{R2}=-\frac{N_{1}\omega_{1}}{N_{2}}.$ (3)

The resulting locking coefficients are presented in Fig. 2 (c) where three
different regimes can be distinguished for the transmission of rotation from
the master to the molecule-gear. In region I (white) for $\tau_{ext}<18$
nN$\cdot$Å, $|L_{1}|\approx|L_{2}|$. In this case, the average angular
velocity for both the master and the molecule-gear are vanishing. There is no
rotation. It is the so-called underdriving regime[13] . In region II (blue)
and for $18<\tau_{ext}<58$ nN$\cdot$Å, $0<|L_{1}|\approx|L_{2}|<1$. It
corresponds to a transmission of rotation between the master and the molecule-
gear. It is the so-called driving regime. Finally in region III (red) for
$\tau_{ext}>58$ nN$\cdot$Å, we always found $|L_{1}|<|L_{2}|$. As the master
torque is continuing to increase, $|L_{2}|$ is reducing further. This
corresponds to an average angular velocity of the HB-BPB molecule-gear which
cannot follow the master rotation. It is the so-called overdriving regime. Let
us notice that as the torque magnitude continues to increase and becomes much
larger than 68 nN$\cdot$Å, some atoms of the tert-butylphenyl teeth start to
dissociate from the master due to the corresponding extremely high kinetic
energy reached in this condition. Here, the molecule-gear slows down, the
conformation of its teeth is rather deformed and the master rotation started
to be erratic.

###  3.2 Phononic dissipation

In Fig. 1 the main friction requiring the application of a large torque is
coming from the interaction between master and the Cu(111) surface (See Ref.
[16]). To study the phononic friction in action, we design a numerical
experiment only for the master by specifying an initial angular velocity
$\omega_{0}=0.1$ rad/ps for this master without molecular teeth and central
hole for simplicity. For different nanodisk diameters, we let this graphene
nanodisk to progressively slow down and stop due to the friction. With this
initial angular velocity, one can estimate that for atoms away from the
rotational axle the corresponding tangential velocities can exceed 1000 cm/s,
which means that they are in a high-speed friction regime[17]. In this case,
they are subject to viscous dissipation due to the collisions with the surface
Cu(111) atoms which are also displaced because of the master rotation. The
angular velocity decay as a function of time is presented in Fig. 3 for
different nanodisk diameters. For small diameters, the decay time is longer
than for larger diameters because of the large contact area with the Cu(111)
surface for the latter.

Figure 3: Viscous dissipation for graphene disks with initial angular velocity
$\omega_{0}=0.1$ rad/ps on Cu(111) within 200 ps for different gear diameters
ranging from 5 to 15 nm.

To quantify the angular velocity relaxation time, we fit its almost
exponential decay with time by $\omega_{0}e^{-t/\tau}$. The variations of the
inverse of relaxation time or friction coefficient $\eta=1/\tau$ as a function
of the nanodisk diameter are presented in Fig. 4 (blue), for graphene nanodisk
diameters between 3 nm and 20 nm. We found here with MD the friction increases
as size becomes larger but it is weakly depending on size for $d>10$ nm. We
notice that for small diameters ($d<5$ nm), the exponent of $d$ is 2, and this
can be described by the friction theory of isolated adsorbates[18, 19]. The
friction coefficient is given by

$\eta_{\parallel}\approx\frac{3}{8\pi}\frac{M}{\rho}\left(\frac{\omega_{\parallel}}{c_{T}}\right)^{3}\omega_{\parallel},$
(4)

where $M$ is the mass of adsorbate (graphen nanodisk), $\omega_{\parallel}$ is
the resonance frequency for parallel motion, $\rho$ and $c_{T}$ are the mass
density and transverse sound velocity of the substrate. This implies
$\eta\propto M(d)\propto d^{2}$ and is consistent with MD. However, for larger
diameters the increasing of friction coefficient is suppressed since the disk
is approaching an incommensurate layers with respect to substrate and creates
destructive interference[18]. On the other hand, for diameters below 3 nm (red
region in Fig. 3), the nanodisks no longer rotate but oscillate in a Brownian
like motion[17]. The reason for this transition is that the potential energy
barrier height for rotation on the potential energy surface (PES) describing
the rotation mechanics of a graphene nanodisk on a Cu(111) surface becomes in
this case smaller that the initially imposed rotational kinetic energy as
presented in Fig. 5. For example, with a d = 2 nm nanodisk (Fig. 5 (a)), the
total rotational kinetic energy around 35 meV for an initial angular velocity
$\omega_{0}$ = 0.1 rad/ps is below the rotation barrier height of about 50
meV. In this case, the d = 2nm graphene nanodisk oscillates randomly in its
initial potential well considering also the $T$ = 10 K Cu(111) surface
temperature. On the contrary, a d = 3 nm nanodisk has a corresponding kinetic
energy of about 180 meV. Therefore, the rotation of the nanodisk will be
weakly impacted by its mechanical rotation barrier height and its rotational
motion on the Cu(111) surface will be dissipative (Fig. 5 (b)).

Figure 4: Size dependence of rotational friction coefficient $\eta=1/\tau$ due
to the phononic contribution (blue) and the electronic contribution according
to Eq. 5 (red) for disk diameters $d$ from 3 to 20 nm. The red region
indicates that below 3 nm the motion becomes oscillatory due to confinement by
rotational barrier heights. Figure 5: Profiles of potential energy surfaces
$V(\theta)$ and rotational kinetic energies (black line) for (a) 2 nm and (b)
3 nm graphene disk on Cu(111) surface.

###  3.3 Electronic dissipation

Aside from the phononic contribution, it is also important to estimate how
large the electronic contribution is as compared to the phononic one active in
our MD simulations. The electronic friction from the van der Waals interaction
has been studied in Refs.[18, 20] where the metal was treated using a semi-
infinite jellium model. From second-order perturbation theory, one has the
following expression for the electronic friction coefficient (in Gaussian
unit):

$\eta_{el}=1/\tau_{el}=\frac{e^{2}}{\hbar
a_{0}}\frac{\left[k_{F}^{3}\alpha(0)\right]^{2}}{\left(k_{F}z_{0}\right)^{10}}\frac{m_{e}}{M}\frac{\omega_{F}}{\omega_{p}}k_{F}z_{0}I(z_{0},r_{s}),$
(5)

where $e$ is the elementary charge, $\hbar$ is the reduced Planck constant,
$a_{0}$ is the Bohr radius, $\alpha(0)$ the static polarizability of the
adsorbate, $\omega_{p}$ is the plasma frequency of the metallic substrate,
$k_{F}$ the Fermi wave vector, $\omega_{F}$ the Fermi frequency, $m_{e}$ the
electron mass, $M$ the mass of the graphene nanodisk and where $I$ is the
parallel frictional integral (defined in Ref. [21]) as a function of the
electron gas radius $r_{s}$ and of the distance $z_{0}$ between the graphene
nanodisk and the supporting substrate (calculated from MD, this average
distance is $z_{0}=3.1$Å). For Cu(111), $\omega_{p}=1.33\times 10^{16}$
rad/s[22, 23], $\omega_{F}=1.06\times 10^{16}$ rad/s[24], $k_{F}=1.36\times
10^{10}$ A-1[24] and $I(z_{0},r_{s})\approx 7$[21, 18]($r_{s}=2.67$ for Cu).
For monolayer graphene sheet, $\alpha(0)=0.949$ Å3 per unit cell[25]. To
estimate $\alpha(0)$ for a graphene nanodisk, we assume that this disk is
large enough such that $\alpha(0)\approx 0.949\times N_{c}/2$ (one unit cell
contains two carbon atoms) where $N_{c}$ is the number of carbon atoms in the
graphene nanodisk. Then, we can estimate the electronic contribution to the
rotational friction coefficient as a function of the nanodisk diameters. The
results are shown in the red curve in Fig. 4. According to Eq. 5, the
electronic friction coefficient is roughly proportional to $d^{2}$ since the
increase of polarizability is proportional to the number of carbon atoms and
it grows faster than the increase of the mass of a graphene nanodisk
($\alpha(0)^{2}/M\propto d^{2}$).

In comparison with the phononic contribution to friction, the electronic
contribution to friction dominates for larger disks whereas for small size
disks the phononic dissipation dominates. As a consequence a large graphene
nanodisk of higher polarizability could be effective in reducing electronic
friction. For instance, for other metallic substrates like Pb and Nb, the
electronic friction will reduce greatly[26, 27] when $T$ is below the critical
superconducting temperature pointing out that in the set up of Fig. 1 it may
better to employ a superconducting surface support.

##  4 Conclusion and outlook

In conclusion, our large-scale MD simulations demonstrate that a solid state
nanogear equipped with adapted molecular teeth chemical groups will be able to
transfer a rotational motion to a single molecule-gear having the same
molecular teeth. To fight against the surface friction of the supporting
surface, a large torque has to be applied to the solid state nanogear.
Experimentally and at low temperature, such a torque can be applied by one of
the STM tips of a 4 STM tips instrument. But this can destabilize the chemical
structure of the complete machinery in particular disengage the solid state
nanogear from its atomic axle. In this prospect, experiments performed on a
superconducting surface are preferable as recently performed on a molecule-
gear train[6]. We have also assumed that the chemical structure of the solid
state nanodisk is restricted to a one-layer thickness to be compatible with
the effective surface height of the molecule-gear. This puts a large
constraint on the solid state nanogears nanofabrication process whose
thickness in our days are in the 5 nm to 10 nm range. A 2D monolayer material
like graphene or MoS2 is a good starting point for nanofabricating such a
solid state nanogear with a thickness in the 1 nm range which can also be
modulated layer by layer. Finally, we have assumed that both the molecule-gear
and the solid state nanogear are exactly rotating respecting their initial
predefined rotational axles. This hypothesis requires a better exploration but
depends practically on how the central axle of this solid state nanogear with
be atomically nanofabricated. Low temperature STM experiments are now underway
to confirm the above MD simulations.

This work has been supported by the International Max Planck Research School
(IMPRS) for “Many-Particle Systems in Structured Environments”, the MANA-NIMS
MEXT WPI program and also by the European Union Horizon 2020 FET Open project
“Mechanics with Molecules” (MEMO, grant nr. 766864). We also acknowledge the
Center for Information Services and High Performance Computing (ZIH) at TU
Dresden for computational resources.

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##  Appendix A Metallicity variations temperature criterion

Fig. 20 shows the metallicity dependence of the KS relation for a temperature
criterion ($T<10^{3}~{}\rm K$). The dependence on metallicity is slightly
stronger than for a gravitational instability criterion.

Figure 20: Comparison of the resolved (750 pc) neutral and molecular KS
relation at different metallicities (different colours) for the temperature
criterion. The shaded region shows the 16th and 84th percentile scatter for
the solar metallicity simulation. We show the disk averaged KS relation from
K98. The neutral KS relation changes almost two dex in the position of the
break while the molecular KS relation has a change in normalization of more
than two orders of magnitude compared to solar metallicity.


# Novel Approach to Infinite Products Using Multiplicative Modulus Function

C. Ganesa Moorthy  Department of Mathematics
Alagappa University
Karaikudi-630 004, India<EMAIL_ADDRESS>
###### Abstract.

The usual nonnegative modulus function is based on addition. A natural
different modulus function on the set of positive reals is introduced.
Arguments for results for series through the usual modulus function are
transformed to arguments for results for infinite products through the new
modular function for multiplication. Counterparts for Riemann rearrangement
theorem and some tests for convergence are derived. These counterparts are
completely new results and they are different from classical results for
infinite products.

###### Key words and phrases:

Infinite series, Infinite products, Nets.

###### 2010 Mathematics Subject Classification:

Primary 40A20; Secondary 40A05

##  1\. Introduction

After giving the definition of convergence of infinite products of complex
numbers, the following two results can be derived (see, e.g., [1]).
$(\textrm{i})$ For a sequence $(a_{n})_{n=1}^{\infty}$ of positive reals,
$\prod\limits_{n=1}^{\infty}(1+a_{n})$ converges if and only if
$\sum\limits_{n=1}^{\infty}a_{n}$ converges, and if and only if
$\prod\limits_{n=1}^{\infty}(1-a_{n})$ converges. $(\textrm{ii})$ For a
sequence $(a_{n})_{n=1}^{\infty}$ of complex numbers,
$\prod\limits_{n=1}^{\infty}(1+a_{n})$ converges whenever
$\prod\limits_{n=1}^{\infty}(1+|a_{n}|)$ converges. The convergence of
$\prod\limits_{n=1}^{\infty}(1+|a_{n}|)$ in the second result is called
absolute convergence of $\prod\limits_{n=1}^{\infty}(1+a_{n})$. However, our
expectation is that the absolute convergence of
$\prod\limits_{n=1}^{\infty}(1+a_{n})$ should mean the convergence of
$\prod\limits_{n=1}^{\infty}|1+a_{n}|$. This expectation is to be fulfilled by
introducing a natural modulus function for multiplication. The results to be
obtained for infinite products use transformed arguments of arguments used for
results for infinite series. A counterpart of Riemann’s rearrangement theorem
is to be derived. Counterparts of tests for convergence of series are to be
derived for infinite products. Tests for convergence provide necessary
conditions and sufficient conditions for convergence.

There is a natural procedure to convert problems in infinite products to
problems in infinite series. This procedure is the one which uses the
exponential function and the natural logarithmic function for transformation.
For example, convergence of $\prod\limits_{n=1}^{\infty}a_{n}$ is equivalent
to convergence of $\sum\limits_{n=1}^{\infty}\log a_{n}$, and convergence of
$\sum\limits_{n=1}^{\infty}a_{n}$ is equivalent to convergence of
$\prod\limits_{n=1}^{\infty}e^{a_{n}}$, formally. This natural procedure is
not to be applied directly in deriving results for infinite products. However
this transformation is to be applied indirectly in arguments for results to be
derived. Only infinite products with positive factors will be considered. So,
the following simplified definition will be adopted.

######  Definition 1.1.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive reals. Let
$p_{n}=\prod\limits_{k=1}^{n}a_{k}$, for every $n=1,2,3,...$.If
$p_{n}\rightarrow p$ for some real $p>0$ as $n\rightarrow\infty$, then it is
said that the infinite product $\prod\limits_{n=1}^{\infty}a_{n}$ converges to
$p$ and it is written as $p=\prod\limits_{n=1}^{\infty}a_{n}$. Otherwise, it
is said that $\prod\limits_{n=1}^{\infty}a_{n}$ does not converge.

######  Remark 1.2.

The cases $p=0$ and $p=+\infty$ are not included for convergence. A necessary
condition for convergence is $a_{n}\rightarrow 1$ as $n\rightarrow\infty$, but
this is not sufficient. A necessary and sufficient condition is given in the
following known theorem. See [1].

######  Theorem 1.3.

An infinite product $\prod\limits_{n=1}^{\infty}a_{n}$, with $a_{n}>0$,
$\forall n$, converges if and only if for every $\epsilon>0$, there exists an
integer $n_{0}$ such that $|a_{m}a_{m+1}...a_{n}-1|<\epsilon$, whenever
$n>m\geq n_{0}$.

######  Remark 1.4.

Let $1<p(1)<p(2)<$… be a sequence of integers such that $p(n+1)-p(n)\leq M$,
for some finite $M>0$. If
$a_{1}a_{2}...a_{p(1)}a_{p(2)+1}a_{p(2)+2}...a_{p(3)}a_{p(4)+1}...a_{p(2k)+1}...a_{p(2k+1)}$…
and
$a_{p(1)+1}a_{p(1)+2}...a_{p(2)}a_{p(3)+1}a_{p(3)+2}...a_{p(4)}a_{p(5)+1}...a_{p(2k+1)+1}...a_{p(2k+2)}$…converge
to $r$ and $s$, then $a_{1}a_{2}$… converges to $rs$. This result can be
derived by using Theorem 1.3. This can be done by modifying the arguments of
Theorem 8.14 in [1]. For example, if $(b_{n})_{n=1}^{\infty}$ and
$(c_{n})_{n=1}^{\infty}$ are sequences of positive numbers such that
$\prod\limits_{n=1}^{\infty}b_{n}$ and $\prod\limits_{n=1}^{\infty}c_{n}$
converge, then $b_{1}c_{1}b_{2}c_{2}b_{3}c_{3}$… and
$b_{1}c_{1}^{-1}b_{2}c_{2}^{-1}b_{3}c_{3}^{-1}$… converge.

##  2\. Modulus Function for Multiplication:

For a given real number $x$, let

$|x|_{+}=max\\{x,-x\\}=\begin{cases}\ x&\text{if $x\geq 0$}\\\ -x&\text{if
$x\leq 0$}.\end{cases}$

Here $|x|_{+}$ is the usual absolute value $|x|$. Let us call $|$ $|_{+}$ as
additive absolute value function (or additive modulus function). Observe that
$0$ is the additive identity in the additive group of real numbers and $-x$ is
the additive inverse of $x$. Observe also that $1$ is the multiplicative
identity in the multiplicative group of positive numbers, and $x^{-1}$ is the
multiplicative inverse of $x>0$.

######  Definition 2.1.

For a given positive real number $x$, let us define

$|x|_{\times}=max\\{x,x^{-1}\\}=\begin{cases}\ x&\text{if $x\geq 1$}\\\
x^{-1}&\text{if $x\leq 1$}.\end{cases}$

Let us call $|$ $|_{\times}$ as multiplicative absolute value function (or
multiplicative modulus function). Here $|x|_{\times}^{-1}\leq
x\leq|x|_{\times}$. Also, $|xy|_{\times}\leq|x|_{\times}|y|_{\times}$ for
$x>0$ and $y>0$.

Positive and negative parts corresponding to multiplication can also be
defined with an understanding that numbers in $(0,1)$ are multiplicative
negative numbers and that number in $(1,+\infty)$ are multiplicative positive
numbers.

######  Lemma 2.2.

For a given $x>0$, let $p=(|x|_{\times}x)^{1/2}$,
$q=\Big{(}\frac{|x|_{\times}}{x}\Big{)}^{1/2}$ . Then $\frac{p}{q}=x$,
$pq=|x|_{\times}$, $p\geq 1$ and $q\geq 1$.

###### Proof.

Direct verification. ∎

The numbers $p$ and $q$ may be considered as multiplicative positive part and
multiplicative negative part of $x$ in Lemma 2.2.

######  Definition 2.3.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers. Then
$\prod\limits_{n=1}^{\infty}a_{n}$ is said to converge m-absolutely, if
$\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$ converges.

######  Lemma 2.4.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$ converges. Then
$\prod\limits_{n=1}^{\infty}a_{n}$ converges and
$\Big{(}\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}\Big{)}^{-1}\leq\prod\limits_{n=1}^{\infty}a_{n}\leq\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$.

###### Proof.

Note that for $m>n$,

$\displaystyle\prod\limits_{k=n}^{m}\frac{1}{|a_{k}|_{\times}}\leq\prod\limits_{k=n}^{m}a_{k}\leq\prod\limits_{k=n}^{m}|a_{k}|_{\times}.$

By Theorem 1.3, $\prod\limits_{n=1}^{\infty}a_{n}$ converges and

$\displaystyle
0<\Big{(}\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}\Big{)}^{-1}\leq\prod\limits_{n=1}^{\infty}a_{n}\leq\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}<\infty.$

∎

Lemma 2.4 may be roughly stated as “$m$-absolute convergence implies
convergence”.

######  Lemma 2.5.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive reals converging to a
positive number $K$. Let $(t_{n})_{n=1}^{\infty}$ be a sequence of positive
reals such that $\sum\limits_{n=1}^{\infty}t_{n}=+\infty$. Then
$(a_{1}^{t_{1}}a_{2}^{t_{2}}...a_{n}^{t_{n}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}\rightarrow
K$ as $n\rightarrow\infty$.

###### Proof.

Without loss of generality, let us assume that $K=1$. Let $\epsilon>0$ be
given. Find an integer $k$ such that $|a_{n}|_{\times}<(1+\epsilon),$ $\forall
n\geq k$. Find $M>0$ such that $|a_{n}|_{\times}\leq M<\infty$, $\forall n$.
Then

$\displaystyle
1\leq(a_{1}^{t_{1}}a_{2}^{t_{2}}...a_{n}^{t_{n}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}\leq(|a_{1}|_{\times}^{t_{1}}...|a_{k}|_{\times}^{t_{k}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}(|a_{k+1}|_{\times}^{t_{k+1}}...|a_{n}|_{\times}^{t_{n}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}$
$\displaystyle\leq
M^{\frac{t_{1}+...+t_{k}}{t_{1}+t_{2}+...+t_{n}}}(1+\epsilon)^{\frac{t_{k+1}+...+t_{n}}{t_{1}+t_{2}+...+t_{n}}}.$

Then, the right hand side tends to $1+\epsilon$ as $n\rightarrow\infty$, for
every $\epsilon>0$. Thus,
$a_{1}^{t_{1}}a_{2}^{t_{2}}...a_{n}^{t_{n}}\rightarrow 1$ as
$n\rightarrow\infty$. This proves the result. ∎

##  3\. Rearrangements

The first Theorem 3.1 is a transformed version of the classical Riemann
rearrangement theorem. Let us use the word “rearrangement” in the usual sense.

######  Theorem 3.1.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive real numbers such that
$\prod\limits_{n=1}^{\infty}a_{n}$ converges and
$\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$ does not converge. Suppose
$0\leq\alpha\leq\beta\leq\infty$. Then there exists a rearrangement
$\prod\limits_{n=1}^{\infty}a_{n}^{\prime}$ with partial products
$u_{n}^{\prime}=\prod\limits_{k=1}^{n}a_{k}^{\prime}$ such that
$\liminf\limits_{n\rightarrow\infty}u_{n}^{\prime}=\alpha$ and
$\limsup\limits_{n\rightarrow\infty}u_{n}^{\prime}=\beta$.

###### Proof.

Let $p_{n}=(|a_{n}|_{\times}a_{n})^{1/2}$ and
$q_{n}=\Big{(}\frac{|a_{n}|_{\times}}{a_{n}}\Big{)}^{1/2}$, for every $n$. By
Lemma 2.2, $\frac{p_{n}}{q_{n}}=a_{n},p_{n}q_{n}=|a_{n}|_{\times},p_{n}\geq 1$
and $q_{n}\geq 1$, for every $n$. Since
$\prod\limits_{n=1}^{\infty}(p_{n}q_{n})=\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$,
either $\prod\limits_{k=1}^{n}p_{k}\rightarrow+\infty$ as $n\rightarrow\infty$
or $\prod\limits_{k=1}^{n}q_{k}\rightarrow\infty$ as $n\rightarrow\infty$.
Since
$\frac{\prod\limits_{k=1}^{n}p_{k}}{\prod\limits_{k=1}^{n}q_{k}}=\prod\limits_{k=1}^{n}a_{k}$
converges as $n\rightarrow\infty$, if $\prod\limits_{k=1}^{n}p_{k}$ converges
as $n\rightarrow\infty$, then $\prod\limits_{k=1}^{n}q_{k}$ converges as
$n\rightarrow\infty$, and similarly, if $\prod\limits_{k=1}^{n}q_{k}$
converges as $n\rightarrow\infty$, then $\prod\limits_{k=1}^{n}p_{k}$
converges as $n\rightarrow\infty$. Hence
$\prod\limits_{k=1}^{n}p_{k}\rightarrow\infty$ and
$\prod\limits_{k=1}^{n}q_{k}\rightarrow\infty$ as $n\rightarrow\infty$. Now,
let $p_{1}^{\prime},p_{2}^{\prime},...$ denote the factors of
$\prod\limits_{n=1}^{\infty}a_{n}$ which are greater than or equal to $1$ in
the order in which they occur, and let $q_{1}^{\prime},q_{2}^{\prime},...$ be
the multiplicative absolute values of the remaining factors of
$\prod\limits_{n=1}^{\infty}a_{n}$ in their original order. The products
$\prod\limits_{n=1}^{\infty}p_{n}^{\prime}$ and
$\prod\limits_{n=1}^{\infty}q_{n}^{\prime}$ are different from
$\prod\limits_{n=1}^{\infty}p_{n}$ and $\prod\limits_{n=1}^{\infty}q_{n}$ only
by factors $1$, and therefore
$\prod\limits_{k=1}^{n}p_{k}^{\prime}\rightarrow\infty$ and
$\prod\limits_{k=1}^{n}q_{k}^{\prime}\rightarrow\infty$ as
$n\rightarrow\infty$. Choose sequences $(\alpha_{n})_{n=1}^{\infty}$ and
$(\beta_{n})_{n=1}^{\infty}$ of positive numbers such that
$\alpha_{n}\rightarrow\alpha$ and $\beta_{n}\rightarrow\beta$ as
$n\rightarrow\infty,\alpha_{n}<\beta_{n}$, $\forall n$, and $\beta_{1}\geq 1$.
Let $m_{1},k_{1}$ be the smallest integers such that
$p_{1}^{\prime}p_{2}^{\prime}...p_{m_{1}}^{\prime}>\beta_{1}$,
$p_{1}^{\prime}p_{2}^{\prime}...p_{m_{1}}^{\prime}q_{1}^{\prime-1}q_{2}^{\prime-1}...q_{k_{1}}^{\prime-1}<\alpha_{1}$
; let $m_{2},k_{2}$ be the smallest integers such that
$p_{1}^{\prime}p_{2}^{\prime}...p_{m_{1}}^{\prime}q_{1}^{\prime-1}q_{2}^{\prime-1}...q_{k_{1}}^{\prime-1}p_{m_{1}+1}^{\prime}...p_{m_{2}}^{\prime}>\beta_{2}$,
$p_{1}^{\prime}p_{2}^{\prime}...p_{m_{1}}^{\prime}q_{1}^{\prime-1}q_{2}^{\prime-1}...q_{k_{1}}^{\prime-1}p_{m_{1}+1}^{\prime}...p_{m_{2}}^{\prime}q_{k_{1}+1}^{\prime-1}...q_{k_{2}}^{\prime-1}<\alpha_{2}$;
and let us continue in this way. Let $x_{n},y_{n}$ denote the partial products
of the rearrangement
$p_{1}^{\prime}p_{2}^{\prime}...p_{m_{1}}^{\prime}q_{1}^{\prime-1}q_{2}^{\prime-1}...q_{k_{1}}^{\prime-1}p_{m_{1}+1}^{\prime}...p_{m_{2}}^{\prime}q_{k_{1}+1}^{\prime-1}...q_{k_{2}}^{\prime-1}p_{m_{2}+1}^{\prime}...$
whose last factors are $p_{m_{n}}^{\prime},q_{k_{n}}^{\prime-1}$. Then
$\frac{x_{n}}{p_{m_{n}}^{\prime}}\leq\beta_{n}$ and
$\frac{y_{n}}{q_{k_{n}}^{\prime-1}}\geq\alpha_{n}$ so that
$p_{m_{n}}^{\prime-1}\leq\frac{\beta_{n}}{x_{n}}\leq 1$ and
$1\leq\frac{\alpha_{n}}{y_{n}}\leq q_{k_{n}}^{\prime}$. Since
$p_{n}^{\prime}\rightarrow 1$ and $q_{n}^{\prime}\rightarrow 1$ as
$n\rightarrow\infty$, then $x_{n}\rightarrow\beta$ and
$y_{n}\rightarrow\alpha$ as $n\rightarrow\infty$. From our construction, if
$u_{n}^{\prime}$ is the $n$-th partial product of the constructed
rearrangement, then it is clear that
$\liminf\limits_{n\rightarrow\infty}u_{n}^{\prime}=\alpha$ and
$\limsup\limits_{n\rightarrow\infty}u_{n}^{\prime}=\beta$. ∎

######  Remark 3.2.

It should be observed from this Theorem 3.1 that if every rearrangement
converges then $m$-absolute convergence holds.

######  Theorem 3.3.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$ converges. Then every
rearrangement of $\prod\limits_{n=1}^{\infty}a_{n}$ converges and all of them
converge to the same sum.

###### Proof.

Let $M=\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$. Let
$u_{n}=\prod\limits_{k=1}^{n}a_{k}$ and
$u_{n}^{\prime}=\prod\limits_{k=1}^{n}a_{k}^{\prime}$ be partial products of
$\prod\limits_{k=1}^{\infty}a_{k}$ and its one rearrangement
$\prod\limits_{k=1}^{\infty}a_{k}^{\prime}$. To each $n$, let
$A_{n}=\Big{\\{}k\in\\{1,2,...,n\\}:|a_{k}|_{\times}=|a_{k}^{\prime}|_{\times}\Big{\\}}$
and $B_{n}=\\{1,2,...,n\\}\setminus A_{n}$. Then

$\displaystyle
0\leq\Big{|}\prod\limits_{k=1}^{n}|a_{k}|_{\times}-\prod\limits_{k=1}^{n}|a_{k}^{\prime}|_{\times}\Big{|}_{+}\leq
M\Big{|}\prod\limits_{k\in B_{n}}|a_{k}|_{\times}-\prod\limits_{k\in
B_{n}}|a_{k}^{\prime}|_{\times}\Big{|}_{+}$ $\displaystyle\leq
M\Bigg{[}\Big{|}1-\prod\limits_{k\in
B_{n}}|a_{k}|_{\times}\Big{|}_{+}+\Big{|}1-\prod\limits_{k\in
B_{n}}|a_{k}^{\prime}|_{\times}\Big{|}_{+}\Bigg{]}.$

By Theorem 1.3, the right hand side tends to zero as $n\longrightarrow\infty$,
because $|a_{k}|_{\times}\geq 1$ and $|a_{k}^{\prime}|_{\times}\geq 1$,
$\forall k$. Therefore
$\prod\limits_{k=1}^{\infty}|a_{k}|_{\times}=\prod\limits_{k=1}^{\infty}|a_{k}^{\prime}|_{\times}$,
and $\prod\limits_{k=1}^{\infty}|a_{k}^{\prime}|_{\times}$ converges. By Lemma
2.4, $\prod\limits_{k=1}^{\infty}a_{k}^{\prime}$ converges. In the previous
arguments, for $B_{n}$ defined above, let us observe that
$\Big{|}1-\prod\limits_{k\in
B_{n}}|a_{k}|_{\times}^{-1}\Big{|}_{+}+\Big{|}1-\prod\limits_{k\in
B_{n}}|a_{k}^{\prime}|_{\times}\Big{|}_{+}$ tends to zero as
$n\rightarrow\infty$. This is applicable even for subsets of $B_{n}$. So, if
$C_{n}=\Big{\\{}k\in\\{1,2,...,n\\}:a_{k}=a_{k}^{\prime}\Big{\\}}$ and
$D_{n}=\\{1,2,...,n\\}\setminus C_{n}$, then $\Big{|}1-\prod\limits_{k\in
D_{n}}a_{k}\Big{|}_{+}+\Big{|}1-\prod\limits_{k\in
D_{n}}a_{k}^{\prime}\Big{|}_{+}\rightarrow 0$ as $n\rightarrow\infty$, because
$\prod\limits_{k\in D_{n}}|a_{k}|_{\times}^{-1}\leq\prod\limits_{k\in
D_{n}}a_{k}\leq\prod\limits_{k\in D_{n}}|a_{k}|_{\times}$ and
$\prod\limits_{k\in D_{n}}|a_{k}^{\prime}|_{\times}^{-1}\leq\prod\limits_{k\in
D_{n}}a_{k}^{\prime}\leq\prod\limits_{k\in D_{n}}|a_{k}^{\prime}|_{\times}$.
Thus

$\displaystyle
0\leq\Big{|}\prod\limits_{k=1}^{n}a_{k}-\prod\limits_{k=1}^{n}a_{k}^{\prime}\Big{|}_{+}\leq
M\Bigg{[}\Big{|}1-\prod\limits_{k\in
D_{n}}a_{k}\Big{|}_{+}+\Big{|}1-\prod\limits_{k\in
D_{n}}a_{k}^{\prime}\Big{|}_{+}\Bigg{]}$

implies that
$\prod\limits_{k=1}^{\infty}a_{k}=\prod\limits_{k=1}^{\infty}a_{k}^{\prime}$.
∎

######  Definition 3.4.

A product $\prod\limits_{k=1}^{\infty}b_{k}$ is called a subproduct of
$\prod\limits_{k=1}^{\infty}a_{k}$ if $b_{k}=a_{k}$ or $1$ for all $k$, and
$b_{k}=a_{k}$ for infinitely many $k$.

######  Theorem 3.5.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$ converges. Consider a class of
all products of the form $\prod\limits_{n=1}^{\infty}a_{n}^{c_{n}}$, where
$-1\leq c_{n}\leq 1$, $\forall n$. Then all these products converge uniformly
in the following sense. For every $\epsilon>0$, there is an integer $n_{0}$,
which is common for all sequences $(c_{n})_{n=1}^{\infty}$ satisfying $-1\leq
c_{n}\leq 1$, $\forall n$, such that
$\Big{|}1-\prod\limits_{k=n}^{m}a_{k}^{c_{k}}\Big{|}_{+}<\epsilon$, $\forall
m,n\geq n_{0}$ satisfying $m>n$, and so that
$\Big{|}1-\prod\limits_{k=n}^{\infty}a_{k}^{c_{k}}\Big{|}_{+}\leq\epsilon$,
$\forall n\geq n_{0}$.

###### Proof.

$\prod\limits_{k=n}^{m}|a_{k}|_{\times}^{-1}\leq\prod\limits_{k=n}^{m}a_{k}^{c_{k}}\leq\prod\limits_{k=n}^{m}|a_{k}|_{\times}$,
when $-1\leq c_{k}\leq 1$, $\forall k$, and when $m>n$. Thus

$\displaystyle\Big{|}1-\prod\limits_{k=n}^{m}a_{k}^{c_{k}}\Big{|}_{+}\leq\max\Bigg{\\{}\Big{|}1-\prod\limits_{k=n}^{m}|a_{k}|_{\times}^{-1}\Big{|}_{+},\Big{|}1-\prod\limits_{k=n}^{m}|a_{k}|_{\times}\Big{|}_{+}\Bigg{\\}}.$

This observation proves the result, when Theorem 1.3 is applied. ∎

######  Corollary 3.6.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\prod\limits_{n=1}^{\infty}|a_{n}|_{\times}$ converges. Then, all subproducts
$\prod\limits_{k=1}^{\infty}b_{k}$ of $\prod\limits_{k=1}^{\infty}a_{k}$
converge uniformly in the following sense. For every $\epsilon>0$, there is an
integer $n_{0}$, which is common for all subproducts
$\prod\limits_{k=1}^{\infty}b_{k}$, such that
$\Big{|}1-\prod\limits_{k=n}^{m}b_{k}\Big{|}_{+}<\epsilon$, $\forall m,n\geq
n_{0}$ satisfying $m>n$, and so that
$\Big{|}1-\prod\limits_{k=n}^{\infty}b_{k}\Big{|}_{+}\leq\epsilon$, $\forall
n\geq n_{0}$.

Theorem 3.5 is placed in this section, because its proof uses arguments which
are used in the proof of Theorem 3.3. The next section is also devoted to
rearrangements in terms of unordered products.

##  4\. Unordered Products:

Let $I$ be an infinite set. In this section, $E,E_{1},E_{2},F,F_{1},F_{2}$ are
notations to be used for finite subsets of $I$ with or without mentioning
finiteness of them. Let $(a_{i})_{i\in I}$ be a collection of positive
numbers. Let $I_{1}=\\{i\in I:a_{i}>1\\}$, $I_{2}=\\{i\in I:a_{i}<1\\}$ and
$I_{3}=\\{i\in I:a_{i}=1\\}$. These notations are also fixed in this section.
Let us say that $\prod\limits_{i\in I}a_{i}$ converges, if there is a number
$p\in(0,\infty)$ such that for given $\epsilon>0$, there is a finite set
$F\subseteq I$ such that $\Big{|}\prod\limits_{i\in
E}a_{i}-p\Big{|}_{+}<\epsilon$, for every finite set $E$ satisfying
$F\subseteq E\subseteq I$. In this case, let us say that $\prod\limits_{i\in
I}a_{i}$ converges to $p$ and let us write $\prod\limits_{i\in I}a_{i}=p$.
Equivalently, this convergence happens when and only when for given
$\epsilon>0$, there is a set $F\subseteq I$ such that
$\Big{|}\frac{\prod\limits_{i\in E}a_{i}}{p}-1\Big{|}_{+}<\epsilon$, for every
$E$ satisfying $F\subseteq E\subseteq I$. One may find arguments, which are
necessary to prove Theorem 1.3, in the proof of the next lemma.

######  Lemma 4.1.

$\prod\limits_{i\in I}a_{i}$ converges if and only if for given $\epsilon>0$,
there is a finite set $F\subseteq I$ such that $\Big{|}\prod\limits_{i\in
E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever $E\subseteq I\setminus F$.

###### Proof.

Suppose $\prod\limits_{i\in I}a_{i}$ converges to some $p>0$. Find
$F_{1}\subseteq I$ such that $\Big{|}\prod\limits_{i\in
E}a_{i}-p\Big{|}_{+}<\frac{p}{4}$, whenever $F_{1}\subseteq E\subseteq I$, and
so that $0<\frac{3}{4}p<\prod\limits_{i\in E}a_{i}<\frac{5}{4}p$. For given
$\epsilon>0$, there is a set $F_{2}\supseteq F_{1}$ such that
$\Big{|}\prod\limits_{i\in E}a_{i}-\prod\limits_{i\in
F_{2}}a_{i}\Big{|}_{+}<\frac{3}{4}p\epsilon$ whenever $E\supseteq F_{2}$, and
so that $\Big{|}\prod\limits_{i\in E\setminus
F_{2}}a_{i}-1\Big{|}_{+}<\frac{3p\epsilon}{4\prod\limits_{i\in
F_{2}}a_{i}}<\epsilon$. That is, $\Big{|}\prod\limits_{i\in
E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever $E\subseteq I\setminus F_{2}$.

Conversely, assume that for given $\epsilon>0$, there is a set $F$ such that
$\Big{|}\prod\limits_{i\in E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever
$E\subseteq I\setminus F$. In particular, there is a set $F_{1}$ such that
$\Big{|}\prod\limits_{i\in E}a_{i}-1\Big{|}_{+}<\frac{1}{4}$, whenever
$E\subseteq I\setminus F_{1}$, and so that $\frac{3}{4}<\prod\limits_{i\in
E}a_{i}<\frac{5}{4}$. Therefore, $\frac{3}{4}M<\prod\limits_{i\in
E}a_{i}<\frac{5}{4}M$, for any $E\subseteq I$, where $M=\prod\limits_{i\in
F_{1}}a_{i}$. Fix $\epsilon>0$. Find a set $F_{2}\supseteq F_{1}$ such that
$\Big{|}\prod\limits_{i\in E}a_{i}-1\Big{|}_{+}<\frac{3M\epsilon}{4}$,
whenever $E\subseteq I\setminus F_{2}$. Then, for $F_{2}\subseteq
E_{2}\subseteq E_{1}$, $\Big{|}\prod\limits_{i\in
E_{1}}a_{i}-\prod\limits_{i\in
E_{2}}a_{i}\Big{|}_{+}=\Big{|}\prod\limits_{i\in E_{1}\setminus
E_{2}}a_{i}-1\Big{|}_{+}\Big{|}\prod\limits_{i\in
E_{2}}a_{i}\Big{|}^{-1}_{+}<\epsilon$. Therefore, by triangle inequality,
$\Big{|}\prod\limits_{i\in E_{1}}a_{i}-\prod\limits_{i\in
E_{2}}a_{i}\Big{|}_{+}<2\epsilon$, whenever $F_{2}\subseteq E_{1}$ and
$F_{2}\subseteq E_{2}$. Hence $\prod\limits_{i\in I}a_{i}$ converges. ∎

The following inequality will be used at the end of the proof of the next
lemma. $|1-ab|_{+}\leq|1-a|_{+}+|a|_{+}|1-b|_{+}$.

######  Lemma 4.2.

$\prod\limits_{i\in I}a_{i}$ converges if and only if $\prod\limits_{i\in
I_{1}}a_{i}$ converges and $\prod\limits_{i\in I_{2}}a_{i}$ converges.
(Convention: Finite products converge). Moreover, $\prod\limits_{i\in
I}a_{i}=\Big{(}\prod\limits_{i\in I_{1}}a_{i}\Big{)}\Big{(}\prod\limits_{i\in
I_{2}}a_{i}\Big{)}$, in this case.

###### Proof.

Suppose $\prod\limits_{i\in I}a_{i}$ converges. Fix $\epsilon>0$. Then there
is a set $F$ such that $\Big{|}\prod\limits_{i\in
E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever $E\subseteq I\setminus F$. In
particular, $\Big{|}\prod\limits_{i\in E}a_{i}-1\Big{|}_{+}<\epsilon$,
whenever $E\subseteq I_{1}\setminus(I_{1}\cap F)$, and
$\Big{|}\prod\limits_{i\in E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever
$E\subseteq I_{2}\setminus(I_{2}\cap F)$. This proves that $\prod\limits_{i\in
I_{1}}a_{i}$ and $\prod\limits_{i\in I_{2}}a_{i}$ converge. Since
$\prod\limits_{i\in E}a_{i}=\Big{(}\prod\limits_{i\in E\cap
I_{1}}a_{i}\Big{)}\Big{(}\prod\limits_{i\in E\cap I_{2}}a_{i}\Big{)}$ for any
$E$ and multiplication is continuous in the real line, then
$\prod\limits_{i\in I}a_{i}=\Big{(}\prod\limits_{i\in
I_{1}}a_{i}\Big{)}\Big{(}\prod\limits_{i\in I_{2}}a_{i}\Big{)}$. This
continuity of multiplication should be observed in terms of nets over directed
sets $I_{1},I_{2}$, and $I$.

Conversely assume that $\prod\limits_{i\in I_{1}}a_{i}$ and
$\prod\limits_{i\in I_{2}}a_{i}$ converge. Fix $\epsilon>0$. There are
$F_{1}$, $F_{2}$ such that $\Big{|}\prod\limits_{i\in
E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever $E\subseteq I_{1}\setminus F_{1}$,
and $\Big{|}\prod\limits_{i\in E}a_{i}-1\Big{|}_{+}<\epsilon$, whenever
$E\subseteq I_{2}\setminus F_{2}$. Then $\Big{|}\prod\limits_{i\in
E}a_{i}-1\Big{|}_{+}<\epsilon+(1+\epsilon)\epsilon$, whenever $E\subseteq
I\setminus F_{1}\cup F_{2}$. This proves that $\prod\limits_{i\in I}a_{i}$
converges. ∎

######  Remark 4.3.

Suppose $\prod\limits_{i\in I}a_{i}$ converges. For each $n$, let
$G_{n}=\\{i\in I:a_{i}<1-\frac{1}{n}\\}$ and $H_{n}=\\{i\in
I:a_{i}>1+\frac{1}{n}\\}$. Then $G_{n}$ and $H_{n}$ are finite sets, because
$\prod\limits_{i\in I_{1}}a_{i}$ and $\prod\limits_{i\in I_{2}}a_{i}$
converge. So, $I_{1}=\bigcup\limits^{\infty}_{n=1}H_{n}$ and
$I_{2}=\bigcup\limits^{\infty}_{n=1}G_{n}$ are countable sets. Thus, let us
replace $I$ by the countable set $I_{1}\cup I_{2}$, when there is a need for
simplification in notation in connection with convergence. On some occasions,
it will be fixed as $I=\\{1,2,...\\}$. It should be observed that convergence
of all these products should be explained only in terms of nets over directed
sets $I_{1},I_{2}$, and $I$.

######  Theorem 4.4.

The followings are equivalent.
(a) $\prod\limits_{i\in I}|a_{i}|_{\times}$ converges.
(b) $\prod\limits_{i\in I}a_{i}$ converges.
(c) Let $D$ be a directed set which is a cofinal subset of the directed set
$\\{E:E$ is a finite subset of $I\\}$ under the inclusion relation. That is,
if $E$ is any finite subset of $I$, then there is a set $F\in D$ such that
$E\subseteq F$. (See [2] for terminology). Then the net
$\Big{(}\prod\limits_{i\in F}|a_{i}|_{\times}\Big{)}_{F\in D}$ converges for
any $D$ mentioned above. Moreover, all these nets converge to
$\prod\limits_{i\in I}|a_{i}|_{\times}$.
(d) The net $\Big{(}\prod\limits_{i\in F}a_{i}\Big{)}_{F\in D}$ converges for
any $D$ mentioned in (c). Moreover, all these nets converge to
$\prod\limits_{i\in I}a_{i}$.
(e) Consider $I$ in the form $\\{1,2,...\\}$ as it was mentioned in the
previous Remark 4.3, on assuming (a) or (b). Write $I$ in the form
$I=J_{1}\cup J_{2}\cup....$ in which $J_{i}$ are pairwise disjoint infinite
sets. Let us write each $J_{i}$ in the ordered form
$J_{i}=\\{i_{1},i_{2},...\\}$. Then for each $i$, the product
$\prod\limits^{\infty}_{j=1}a_{i_{j}}$ converges, and then the product
$\prod\limits^{\infty}_{i=1}\Big{(}\prod\limits^{\infty}_{j=1}a_{i_{j}}\Big{)}$
converges, for any decomposition $J_{1}\cup J_{2}\cup....$ of $I$. In this
case, they all converge to $\prod\limits_{i\in I}a_{i}$.
Note: The statement in part (e) is given in this specific form only for the
purpose of the implication: (a) or (b) implies (e). For the reverse
implication, let us take I as a countable set.

###### Proof.

Suppose $\prod\limits_{i\in I}|a_{i}|_{\times}$ converges. Then for any finite
subset $F$ of $I$,

$\displaystyle\Big{(}\prod\limits_{i\in
F}|a_{i}|_{\times}\Big{)}^{-1}\leq\prod\limits_{i\in
F}a_{i}\leq\prod\limits_{i\in F}|a_{i}|_{\times},$

and Lemma 4.1 implies that $\prod\limits_{i\in I}a_{i}$ converges.

Conversely assume that $\prod\limits_{i\in I}a_{i}$ converges. Then
$\prod\limits_{i\in I_{1}}a_{i}$ and $\prod\limits_{i\in I_{2}}a_{i}$
converges, by Lemma 4.2. So, $\prod\limits_{i\in I_{1}}|a_{i}|_{\times}$ and
$\prod\limits_{i\in I_{2}}|a_{i}|_{\times}$ converges. By continuity of
multiplication, $\prod\limits_{i\in I}|a_{i}|_{\times}$ converges.

Thus (a) and (b) are equivalent. From the definition for net convergence, (a),
(b), (c), (d) are equivalent, and common limits for nets also do exist for (c)
and (d).

To prove the equivalence of (a) and (e), suppose that $\prod\limits_{i\in
I}|a_{i}|_{\times}$ converges. By Lemma 4.1, and Theorem 1.3,
$\prod\limits^{\infty}_{j=1}|a_{i_{j}}|_{\times}$ converges for each $i$, when
the notations given in (e) are used. Moreover,

$\displaystyle
1\leq\prod\limits^{n}_{i=m}\prod\limits^{\infty}_{j=1}|a_{i_{j}}|_{\times}\leq\prod\limits_{i\in
I}|a_{i}|_{\times},$

for any $n,m$ satisfying $n>m$. Again by Lemma 4.1 and Theorem 1.3, it can be
observed that
$\prod\limits^{\infty}_{i=1}\Big{(}\prod\limits^{\infty}_{j=1}|a_{i_{j}}|_{\times}\Big{)}$
converges. The inequalities, for $n>m$,

$\displaystyle\Big{(}\prod\limits^{n}_{j=m}|a_{i_{j}}|_{\times}\Big{)}^{-1}\leq\prod\limits^{n}_{j=m}a_{i_{j}}\leq\prod\limits^{n}_{j=m}|a_{i_{j}}|_{\times}$

give the convergence of $\prod\limits^{\infty}_{j=1}a_{i_{j}}$, by Theorem
1.3. Similarly, the inequalities, for $n>m$,

$\displaystyle\Big{(}\prod\limits^{n}_{i=m}\Big{(}\prod\limits^{\infty}_{j=1}|a_{i_{j}}|_{\times}\Big{)}\Big{)}^{-1}\leq\prod\limits^{n}_{i=m}\prod\limits^{\infty}_{j=1}a_{i_{j}}\leq\prod\limits^{n}_{i=m}\prod\limits^{\infty}_{j=1}|a_{i_{j}}|_{\times}$

give the convergence of
$\prod\limits^{\infty}_{i=1}\prod\limits^{\infty}_{j=1}a_{i_{j}}$.

To prove the converse part, assume that $\prod\limits_{i\in
I}|a_{i}|_{\times}$ does not converge, or equivalently
$\prod\limits^{\infty}_{n=1}|a_{n}|_{\times}$ does not converge, when $I$ is
considered in the form $\\{1,2,...\\}=I_{1}\cup I_{2}$ as stated in the
statement for (e). Then either $\prod\limits_{i\in
I_{1}}|a_{i}|_{\times}=+\infty$ or $\prod\limits_{i\in
I_{2}}|a_{i}|_{\times}=+\infty$. For example, let us consider the case
$\prod\limits_{i\in I_{1}}|a_{i}|_{\times}=+\infty$. In this case, let us find
an infinite subset $J_{1}=\\{1_{1},1_{2},...\\}$ of $I$ such that
$I_{1}\setminus J_{1}$ is infinite and such that
$\prod\limits^{\infty}_{j=1}a_{1_{j}}$ does not converge. This is possible in
view of Remark 3.2 and Remark 1.4. Write $(I_{1}\setminus J_{1})\cup I_{2}$ in
the form $J_{2}\cup J_{2}\cup...$ such that each $J_{i}=\\{i_{1},i_{2},...\\}$
is an infinite set and such that $J_{i}$ are pairwise disjoint. Then
$\prod\limits^{\infty}_{i=1}\prod\limits^{\infty}_{j=1}a_{i_{j}}$ does not
converge.

It remains to establish the final part of (e). Let
$p=\prod\limits^{\infty}_{i=1}\prod\limits^{\infty}_{j=1}a_{i_{j}}$ for some
partition $I=J_{1}\cup J_{2}\cup...$. For each $n$, it is possible to find
integers $m(n),n_{1},n_{2},...,n_{m(n)}$ such that
$\Big{|}\prod\limits^{m(n)}_{i=1}\prod\limits^{n_{i}}_{j=1}a_{i_{j}}-p\Big{|}_{+}<\frac{1}{n}$,
and such that $k_{1>}n_{1},k_{2}>n_{2},...,k_{m(n)}>n_{m(n)}$, $m(k)>m(n)$,
whenever $k>n$. Let $D=\Big{\\{}\\{i_{j}:1\leq j\leq n_{i},1\leq i\leq
m(n)\\}:n=1,2,...\Big{\\}}$. Consider $D$ as a directed set under inclusion
relation, and when $I$ is considered as $\\{i_{j}:j=1,2,...,i=1,2,...\\}$. Now
the final part of (d) implies that $p=\prod\limits_{i\in I}a_{i}$. The proof
is now completed. ∎

######  Remark 4.5.

In the previous Theorem 4.4, (e) includes iterated convergence for double
products. The part (d) includes some special types of net convergence meant
for double products corresponding to the one known for double series in
general forms. References for such general forms may be found in [3]. By
Theorem 3.3 and Remark 3.2, it can be stated that all parts (a), (b), (c),
(d),(e) are equivalent to one more part. (f): Consider $I$ in the form
$\\{1,2,...\\}$ (as it was done in (e)). Then all rearrangements
$\prod\limits^{\infty}_{n=1}a_{n}$ of $\prod\limits_{i\in I}a_{i}$ converge.
Moreover, they all converge to $\prod\limits_{i\in I}a_{i}$.

##  5\. Tests For Convergence

If a necessary condition for convergence is not satisfied then convergence
fails. If a sufficient condition for convergence is satisfied then convergence
is assured. All results giving conditions for convergence are classified under
tests for convergence. For example, if
$\limsup\limits_{n\rightarrow\infty}|\log a_{n}|_{+}^{\frac{1}{n}}<1$ then
$\prod\limits^{\infty}_{n=1}a_{n}$ converges, formally, according to the root
test for series convergence. This statement can be made into a logically
correct statement. But, it is not our aim. Our aim is to use necessary
transformations in arguments of proofs of results. Let us consider again
countable infinite products.

######  Proposition 5.1.

Let $(t_{n})_{n=1}^{\infty}$ and $(a_{n})_{n=1}^{\infty}$ be sequences of
positive numbers such that $\sum\limits^{\infty}_{n=1}t_{n}$ converges and
such that
$\limsup\limits_{n\rightarrow\infty}a_{n}^{\frac{1}{t_{n}}}=\beta<\infty$ and
$\liminf\limits_{n\rightarrow\infty}a_{n}^{\frac{1}{t_{n}}}=\alpha>0$. Then
$\prod\limits^{\infty}_{n=1}a_{n}$ converges.

###### Proof.

Fix $\alpha^{{}^{\prime}}$ and $\beta^{{}^{\prime}}$ such that
$0<\alpha^{{}^{\prime}}<\alpha\leq\beta<\beta^{{}^{\prime}}<\infty$. Then
there is an integer $n_{0}$ such that
$\alpha^{{}^{\prime}}<a_{n}^{\frac{1}{t_{n}}}<\beta^{{}^{\prime}}$, $\forall$
$n\geq n_{0}$. Then, for $n>m\geq n_{0}$,

$\displaystyle{\alpha^{{}^{\prime}}}^{\sum\limits^{n}_{k=m}t_{k}}<\prod\limits^{n}_{k=m}a_{n}<{\beta^{{}^{\prime}}}^{\sum\limits^{n}_{k=m}t_{k}}.$

Now, the proposition follows from Theorem 1.3. ∎

######  Corollary 5.2.

Let $(t_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\sum\limits^{\infty}_{n=1}t_{n}$ converges. Let $(a_{n})_{n=1}^{\infty}$ be a
sequence of positive numbers such that $a_{n}\geq 1$, $\forall n$, and such
that $\limsup\limits_{n\rightarrow\infty}a_{n}^{\frac{1}{t_{n}}}<\infty$. Then
$\prod\limits^{\infty}_{n=1}a_{n}$ converges. Here $``a_{n}\geq 1"$ and
$``\limsup\limits_{n\rightarrow\infty}a_{n}^{\frac{1}{t_{n}}}<\infty"$ can be
replaced by $``a_{n}\leq 1"$ and
$``\liminf\limits_{n\rightarrow\infty}a_{n}^{\frac{1}{t_{n}}}>0"$ to get the
same conclusion: $\prod\limits^{\infty}_{n=1}a_{n}$ converges.

######  Theorem 5.3.

Suppose $a_{1}\geq a_{2}\geq...\geq 1$.Then $\prod\limits^{\infty}_{n=1}a_{n}$
converges if and only if
$\prod\limits^{\infty}_{k=0}a_{2^{k}}^{2^{k}}=a_{1}^{1}a_{2}^{2}a_{4}^{4}a_{8}^{8}....$
converges.

###### Proof.

Let $u_{n}=a_{1}a_{2}...a_{n}$, and
$v_{k}=a_{2^{0}}^{2^{0}}a_{2^{1}}^{2^{1}}a_{2^{2}}^{2^{2}}....a_{2^{k}}^{2^{k}}$.
For $n<2^{k}$,

$\displaystyle u_{n}\leq
a_{1}(a_{2}a_{3})...(a_{2^{k}}a_{2^{k}+1}...a_{2^{k+1}-1})\leq
a_{1}^{1}a_{2}^{2}a_{4}^{4}...a_{2^{k}}^{2^{k}}=v_{k}$

so that $u_{n}\leq v_{k}$. On the other hand, if $n>2^{k}$,

$\displaystyle u_{n}\geq
a_{1}^{\frac{1}{2}}a_{2}(a_{3}a_{4})...(a_{2^{k-1}-1}...a_{2^{k}})\geq(a_{1}^{1}a_{2}^{2}a_{4}^{4}...a_{2^{k}}^{2^{k}})^{\frac{1}{2}}=v_{k}^{\frac{1}{2}}$

so that $u_{n}^{2}\geq v_{k}$. These two inequalities establish the result,
because $a_{n}\geq 1$, $\forall n$. ∎

######  Definition 5.4.

Suppose $a_{n}>1$, for each $n$. The product
$a_{1}a_{2}^{-1}a_{3}a_{4}^{-1}a_{5}a_{6}^{-1}$…. is called an alternating
product.

######  Theorem 5.5.

If $(a_{n})_{n=1}^{\infty}$ is a strictly decreasing sequence converging to 1,
then the alternating product $a_{1}a_{2}^{-1}a_{3}a_{4}^{-1}a_{5}a_{6}^{-1}$…
converges.

###### Proof.

Let $b_{1}=a_{1}a_{2}^{-1},b_{2}=a_{3}a_{4}^{-1},b_{3}=a_{5}a_{6}^{-1}$,…
.Then $b_{n}>1$, $\forall n$. Hence, $\prod\limits^{\infty}_{n=1}b_{n}$
converges if and only if $(\prod\limits^{n}_{k=1}b_{k})^{\infty}_{n=1}$ is a
bounded sequence. Let us observe that
$\prod\limits^{n}_{k=1}b_{k}=a_{1}(\frac{a_{2}}{a_{3}})^{-1}(\frac{a_{4}}{a_{5}})^{-1}...$
$(\frac{a_{2n-2}}{a_{2n-1}})^{-1}a_{2n}^{-1}<a_{1}$, because
$(\frac{a_{2}}{a_{3}})^{-1},(\frac{a_{4}}{a_{5}})^{-1},...(\frac{a_{2n-1}}{a_{2n-1}})^{-1},a_{2n}^{-1}$
are less than 1. So, $\prod\limits^{\infty}_{n=1}b_{n}$ converges. The value
of $a_{1}a_{2}^{-1}a_{3}a_{4}^{-1}a_{5}a_{6}^{-1}...a_{m}^{(-1)^{m+1}}$ is
$\prod\limits^{\frac{m}{2}}_{k=1}b_{k}$ if $m$ is even, and it is
$\Big{(}\prod\limits^{\frac{m-1}{2}}_{k=1}b_{k}\Big{)}a_{m}$ if $m$ is odd.
So, $a_{1}a_{2}^{-1}a_{3}a_{4}^{-1}$… converges to
$\prod\limits^{\infty}_{n=1}b_{n}$, because $a_{m}\rightarrow 1$ as
$m\rightarrow\infty$. ∎

######  Corollary 5.6.

If $(a_{n})_{n=1}^{\infty}$ is a strictly decreasing sequence in $(0,1)$ which
converges to zero, then $(1-a_{1})(1-a_{2})^{-1}(1-a_{3})(1-a_{4})^{-1}$…
converges.

###### Proof.

Here, $((1-a_{n})^{-1})_{n=1}^{\infty}$ is a strictly decreasing sequence,
which converges to 1. ∎

######  Theorem 5.7.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\prod\limits^{\infty}_{n=1}a_{n}$ converges to $p$. Let
$(t_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$t_{1}+t_{2}+...=+\infty$. Let $u_{n}=\prod\limits^{n}_{k=1}a_{k}$, and
$\sigma_{n}=(u_{1}^{t_{1}}u_{2}^{t_{2}}...u_{n}^{t_{n}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}$,
$\forall n$. Then $\sigma_{n}$ converges to $p$ as $n\rightarrow\infty$.

###### Proof.

Let $v_{n}=\frac{u_{n}}{p}$ and $\tau_{n}=\frac{\sigma_{n}}{p}$, $\forall n$.
Then
$(v_{1}^{t_{1}}v_{2}^{t_{2}}...v_{n}^{t_{n}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}=\frac{\sigma_{n}}{p}$,
$\forall n$. Choose a finite positive number A such that $|v_{n}|_{\times}\leq
A$, $\forall n$. Given $\epsilon>0$, choose an integer $k$ such that
$|v_{n}|_{\times}\leq 1+\epsilon$, $\forall n>k$. Then

$\displaystyle
1\leq|\tau_{n}|_{\times}=\Big{|}\frac{\sigma_{n}}{p}\Big{|}_{\times}\leq(|v_{1}|_{\times}^{t_{1}}|v_{2}|_{\times}^{t_{2}}...|v_{k}|_{\times}^{t_{k}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}(|v_{k+1}|_{\times}^{t_{k+1}}...|v_{n}|_{\times}^{t_{n}})^{\frac{1}{t_{1}+t_{2}+...+t_{n}}}$
$\displaystyle\leq
A^{\frac{t_{1}+t_{2}+...+t_{k}}{t_{1}+t_{2}+...+t_{n}}}(1+\epsilon)^{\frac{t_{k+1}+t_{2}+...+t_{n}}{t_{1}+t_{2}+...+t_{n}}}.$

Thus $1\leq|\tau_{n}|_{\times}\leq 1+2\epsilon$ for sufficiently large $n$,
because $t_{1}+t_{2}+...=\infty$. This proves that
$\lim\limits_{n\rightarrow\infty}|\tau_{n}|_{\times}=1=\lim\limits_{n\rightarrow\infty}\tau_{n}$.
So, $\lim\limits_{n\rightarrow\infty}\sigma_{n}=p.$ ∎

######  Corollary 5.8.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive numbers such that
$\prod\limits^{\infty}_{n=1}a_{n}$ converges to $p$. Let
$u_{n}=\prod\limits^{n}_{k=1}a_{k}$, and
$\sigma_{n}=(u_{1}u_{2}...u_{n})^{\frac{1}{n}}$, $\forall n$. Then
$\sigma_{n}$ converges to $p$ as $n\rightarrow\infty$.

######  Theorem 5.9.

Suppose $(b_{n})_{n=1}^{\infty}$ be a sequence of positive reals such that
$\prod\limits^{\infty}_{n=1}\frac{b_{n}}{b_{n+1}}$ converges m-absolutely and
such that $b_{n}\rightarrow 1$ as $n\rightarrow\infty$. Suppose
$(a_{n})_{n=1}^{\infty}$ be a sequence of real numbers such that
$\Big{\\{}\sum\limits^{m}_{k=n}a_{k}:m=1,2,...,n=1,2,...,n\leq m\Big{\\}}$ is
a bounded set. Then $\prod\limits^{\infty}_{k=1}b_{k}^{a_{k}}$ converges.

###### Proof.

Suppose $\Big{|}\sum\limits^{m}_{k=n}a_{k}\Big{|}_{+}\leq M$, $\forall$
$n=1,2,...$, $\forall m=1,2,...,$ subject to the condition $n\leq m$, for some
$M>0$. Note that, for $n>2$,

$\displaystyle
b_{1}^{a_{1}}b_{2}^{a_{2}}...b_{n}^{a_{n}}=\Big{(}\frac{b_{1}}{b_{2}}\Big{)}^{a_{1}}\Big{(}\frac{b_{2}}{b_{3}}\Big{)}^{a_{1}+a_{2}}...\Big{(}\frac{b_{n-1}}{b_{n}}\Big{)}^{a_{1}+a_{2}+...+a_{n-1}}b_{n}^{a_{1}+a_{2}+...+a_{n}}$

Then

$\displaystyle\Big{(}\Big{|}\frac{b_{1}}{b_{2}}\Big{|}_{\times}\Big{|}\frac{b_{2}}{b_{3}}\Big{|}_{\times}...\Big{|}\frac{b_{n-1}}{b_{n}}\Big{|}_{\times}\Big{)}^{-M}|b_{n}|_{\times}^{-M}\leq
b_{1}^{a_{1}}b_{2}^{a_{2}}...b_{n}^{a_{n}}\leq\Big{(}\Big{|}\frac{b_{1}}{b_{2}}\Big{|}_{\times}\Big{|}\frac{b_{2}}{b_{3}}\Big{|}_{\times}...\Big{|}\frac{b_{n-1}}{b_{n}}\Big{|}_{\times}\Big{)}^{M}|b_{n}|_{\times}^{M}.$

One can derive similar inequalities for
$b_{m}^{a_{m}}b_{m+1}^{a_{m+1}}...b_{n}^{a_{n}}$, and for $m<n$, and then
Theorem 1.3 can be applied to derive convergence of
$\prod\limits^{\infty}_{k=1}{b_{k}}^{a_{k}}$. ∎

This Theorem 5.9 gives a motivation for concepts and results to be given in
Sections 6 and 7. Theorem 5.7 and Theorem 5.9 of this Section 5 are comparable
with the corresponding classical summability results, Theorem 8.48 in [1] and
Theorem 8.27 in [1], which attributed to the mathematicians Cesaro and Abel,
respectively.

##  6\. Matrices and Multiplicability

Let $(0,\infty)^{n}$ denote the cartesian product
$\prod\limits^{n}_{i=1}X_{i}$ in which each $X_{i}=(0,\infty)$, multiplication
is defined coordinatewisely, and elements are written as transpose of row
vectors $(x_{1},x_{2},...,x_{n})$. More explicitly,

$\begin{pmatrix}x_{1}\\\ x_{2}\\\ .\\\ .\\\ .\\\ x_{n}\\\
\end{pmatrix}\begin{pmatrix}y_{1}\\\ y_{2}\\\ .\\\ .\\\ .\\\ y_{n}\\\
\end{pmatrix}=\begin{pmatrix}x_{1}y_{1}\\\ x_{2}y_{2}\\\ .\\\ .\\\ .\\\
x_{n}y_{n}\\\ \end{pmatrix}$

or
$(x_{1},x_{2},...,x_{n})^{T}(y_{1},y_{2},...,y_{n})^{T}=(x_{1}y_{1},x_{2}y_{2},...,x_{n}y_{n})^{T}$
in $(0,\infty)^{n}$. Let $A=(a_{ij})$ be a matrix of order $m\times n$ with
real entries $a_{ij}$. Let us define

$A*\begin{pmatrix}x_{1}\\\ x_{2}\\\ .\\\ .\\\ .\\\ x_{n}\\\
\end{pmatrix}=\begin{pmatrix}x_{1}^{a_{11}}x_{2}^{a_{12}}...x_{n}^{a_{1n}}\\\
x_{1}^{a_{21}}x_{2}^{a_{22}}...x_{n}^{a_{2n}}\\\ .\\\ .\\\ .\\\
x_{1}^{a_{m1}}x_{2}^{a_{m1}}...x_{n}^{a_{mn}}\\\ \end{pmatrix}$

such that A is a multiplication preserving function from $(0,\infty)^{n}$ to
$(0,\infty)^{m}$. Then, for a given matrix $B=(b_{ij})$ of order $k\times m$
with real entries, it can be verified that

$B*\begin{pmatrix}A*\begin{pmatrix}x_{1}\\\ x_{2}\\\ .\\\ .\\\ .\\\ x_{n}\\\
\end{pmatrix}\end{pmatrix}=(BA)*\begin{pmatrix}x_{1}\\\ x_{2}\\\ .\\\ .\\\
.\\\ x_{n}\\\ \end{pmatrix},$

$\forall$ $(x_{1},x_{2},...,x_{n})^{T}\in(0,\infty)^{n}$, where $BA$ is the
usual matrix multiplication. Let $C$ be another matrix of order $k\times m$
with real entries. Then the following is true for every
$(x_{1},x_{2},...,x_{n})^{T}\in(0,\infty)^{n}$:

$(C+B)*\begin{pmatrix}A*\begin{pmatrix}x_{1}\\\ x_{2}\\\ .\\\ .\\\ .\\\
x_{n}\\\
\end{pmatrix}\end{pmatrix}=\begin{pmatrix}C*\begin{pmatrix}A*\begin{pmatrix}x_{1}\\\
x_{2}\\\ .\\\ .\\\ .\\\ x_{n}\\\
\end{pmatrix}\end{pmatrix}\end{pmatrix}\begin{pmatrix}B*\begin{pmatrix}A*\begin{pmatrix}x_{1}\\\
x_{2}\\\ .\\\ .\\\ .\\\ x_{n}\\\ \end{pmatrix}\end{pmatrix}\end{pmatrix}.$

Also,

$(C+B)*\begin{pmatrix}y_{1}\\\ y_{2}\\\ .\\\ .\\\ .\\\ y_{m}\\\
\end{pmatrix}=\begin{pmatrix}C*\begin{pmatrix}y_{1}\\\ y_{2}\\\ .\\\ .\\\ .\\\
y_{m}\\\ \end{pmatrix}\end{pmatrix}\begin{pmatrix}B*\begin{pmatrix}y_{1}\\\
y_{2}\\\ .\\\ .\\\ .\\\ y_{m}\\\ \end{pmatrix}\end{pmatrix},$

$\forall$ $(y_{1},y_{2},...,y_{m})^{T}\in(0,\infty)^{m}$. Let $D$ and $E$ be
matrices of order $n\times k$ with real entries. Then it can be verified that

$A*\begin{pmatrix}(D+E)*\begin{pmatrix}z_{1}\\\ z_{2}\\\ .\\\ .\\\ .\\\
z_{k}\\\
\end{pmatrix}\end{pmatrix}=\begin{pmatrix}A*\begin{pmatrix}D*\begin{pmatrix}z_{1}\\\
z_{2}\\\ .\\\ .\\\ .\\\ z_{k}\\\
\end{pmatrix}\end{pmatrix}\end{pmatrix}\begin{pmatrix}A*\begin{pmatrix}E*\begin{pmatrix}z_{1}\\\
z_{2}\\\ .\\\ .\\\ .\\\ z_{k}\\\ \end{pmatrix}\end{pmatrix}\end{pmatrix}$

$\forall$ $(z_{1},z_{2},...,z_{k})^{T}\in(0,\infty)^{k}$. Thus, there are
interesting matrix operations. With this familiarity of the operation $*$ for
finite matrices, let us extend the same for infinite matrices or double
sequences which are associated with summability. Corollary 5.8 is also a
counterpart of a summability method.

######  Proposition 6.1.

Let $(a_{m,n})_{m,n=1}^{\infty}$ be a double sequence of positive numbers such
that $\sum\limits^{\infty}_{n=1}a_{m,n}\leq M$, for every $m$, for some $M>0$,
and such that $\lim\limits_{m\rightarrow\infty}a_{m,n}=0$, for every $n$.
Then, for every sequence $(x_{n})_{n=1}^{\infty}$ of positive reals for which
$\lim\limits_{n\rightarrow\infty}x_{n}=p>0$ exists,
$\lim\limits_{m\rightarrow\infty}\prod\limits^{\infty}_{n=1}(\frac{x_{n}}{p})^{a_{m,n}}=1$.
Equivalently,
$\lim\limits_{m\rightarrow\infty}\prod\limits^{\infty}_{n=1}{x_{n}}^{a_{m,n}}=1$,
whenever $\lim\limits_{n\rightarrow\infty}x_{n}=1$ and $x_{n}>0$, $\forall n$.

###### Proof.

Suppose $(x_{n})_{n=1}^{\infty}$ be a sequence of positive reals such that
$x_{n}\rightarrow 1$ as $n\rightarrow\infty$. Fix $\epsilon>0$. Find an
integer $n_{0}$ such that $|x_{k}|_{\times}<(1+\epsilon)^{\frac{1}{2M}}$,
$\forall$ $k\geq n_{0}$. Then, for $l>k\geq n_{0}$,

$\displaystyle(1+\epsilon)^{\frac{-1}{2}}<\prod\limits^{l}_{s=k}|x_{s}|_{\times}^{-a_{m,s}}\leq\prod\limits^{l}_{s=k}{x_{s}}^{a_{m,s}}\leq\prod\limits^{l}_{s=k}|x_{s}|_{\times}^{a_{m,s}}<(1+\epsilon)^{\frac{1}{2}},$

$\forall$ $m$, and hence

$\displaystyle(1+\epsilon)^{\frac{-1}{2}}\leq\prod\limits^{\infty}_{n=n_{0}+1}{x_{n}}^{a_{m,n}}\leq(1+\epsilon)^{\frac{1}{2}},$

$\forall$ $m$. Find an integer $m_{0}$ such that

$\displaystyle(1+\epsilon)^{\frac{-1}{2}}<\prod\limits^{n_{0}}_{k=1}|x_{k}|_{\times}^{-a_{m,k}}\leq\prod\limits^{n_{0}}_{k=1}{x_{k}}^{a_{m,k}}\leq\prod\limits^{n_{0}}_{k=1}|x_{k}|_{\times}^{a_{m,k}}<(1+\epsilon)^{\frac{1}{2}},$

$\forall$ $m\geq m_{0}$. Thus, for $m\geq m_{0}$,

$\displaystyle(1+\epsilon)^{-1}\leq\prod\limits^{n_{0}}_{n=1}{x_{n}}^{a_{m,n}}\prod\limits^{\infty}_{n=n_{0}+1}{x_{n}}^{a_{m,n}}\leq(1+\epsilon).$

That is, for given $\epsilon>0$, there is an integer $m_{0}$ such that

$\displaystyle(1+\epsilon)^{-1}\leq\prod\limits^{\infty}_{n=1}{x_{n}}^{a_{m,n}}\leq(1+\epsilon),$

$\forall$ $m\geq m_{0}$. This proves that
$\lim\limits_{m\rightarrow\infty}\prod\limits^{\infty}_{n=1}{x_{n}}^{a_{m,n}}=1$.
∎

##  7\. Power Products

Corresponding to power series, it is possible to introduce power products.

######  Definition 7.1.

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive reals. A formal power
product is $a_{0}{a_{1}}^{x}{a_{2}}^{x^{2}}{a_{3}}^{x^{3}}$… or
$\prod\limits^{\infty}_{n=0}{a_{n}}^{x^{n}}$.

######  Remark 7.2.

This formal power product converges to a value, if
$|x|_{+}<\limsup\limits_{n\rightarrow\infty}|\log a_{n}|_{+}^{\frac{1}{n}}$.
This can be derived from the corresponding power series obtained by
logarithmic transformation. This is not our method of deriving results. Let us
now recall Theorem 5.9 in another version.

######  Theorem 7.3.

Suppose $(b_{n})_{n=1}^{\infty}$ be as in Theorem 5.9. Then the power product
$\prod\limits^{\infty}_{k=1}{b_{k}}^{x^{k}}$ converges for every real
$x\in[-1,1)$.

###### Proof.

Let $x\in[-1,1)$. Write $a_{k}=x^{k}$, for $k=1,2,...$. Then
$\Big{\\{}\sum\limits^{m}_{k=n}a_{k}:m=1,2,...,n=1,2,...,n\leq m\Big{\\}}$ is
a bounded set. By Theorem 5.9, $\prod\limits^{\infty}_{k=1}{b_{k}}^{x^{k}}$
converges. ∎

######  Corollary 7.4.

Let $(b_{n})_{n=1}^{\infty}$ be a strictly increasing sequence of positive
reals converging to 1. Then the power product
$\prod\limits^{\infty}_{k=1}{b_{k}}^{x^{k}}$ converges for every real
$x\in[-1,1)$.

###### Proof.

For $n>m$,

$\displaystyle 1$ $\displaystyle\leq$
$\displaystyle\Big{|}\frac{b_{m}}{b_{m+1}}\Big{|}_{\times}\Big{|}\frac{b_{m+1}}{b_{m+2}}\Big{|}_{\times}...\Big{|}\frac{b_{n}}{b_{n+1}}\Big{|}_{\times}$
$\displaystyle=$
$\displaystyle\frac{b_{m+1}}{b_{m}}\frac{b_{m+2}}{b_{m+1}}...\frac{b_{n+1}}{b_{n}}=\frac{b_{n+1}}{b_{m}}<\frac{1}{b_{m}},$

and $\frac{1}{b_{m}}\rightarrow 1$ as $m\rightarrow\infty$. Therefore, by
Theorem 1.3, $\prod\limits^{\infty}_{n=1}|\frac{b_{n}}{b_{n+1}}|_{\times}$
converges. Now, the conclusion follows from Theorem 7.3. ∎

So, there is a need to improve Theorem 5.9. The next result may remind Cauchy
product for sequences.

######  Theorem 7.5.

Let $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$ be two sequences of
positive reals such that $\prod\limits^{\infty}_{n=0}a_{n}$ and
$\prod\limits^{\infty}_{n=0}b_{n}$ converge to $A$ and $B$. For each $n$, let
$c_{n}=a_{0}b_{n}a_{1}b_{n-1}...a_{n}b_{0}$,
$A_{n}=\prod\limits^{n}_{k=0}a_{k}$, $B_{n}=\prod\limits^{n}_{k=0}b_{k}$ and
$C_{n}=\prod\limits^{n}_{k=0}c_{k}$. Then ${A_{n}}^{\frac{1}{n}}\rightarrow
1$, ${B_{n}}^{\frac{1}{n}}\rightarrow 1$, ${C_{n}}^{\frac{1}{n}}\rightarrow
AB$ and ${c_{n}}^{\frac{1}{n}}\rightarrow 1$ as $n\rightarrow\infty$.

###### Proof.

By Theorem 1.3, $a_{n}\rightarrow 1$, $b_{n}\rightarrow 1$ as
$n\rightarrow\infty$. By Lemma 2.5, all conclusions follow, because
$C_{n}=A_{0}A_{1}...A_{n}B_{0}B_{1}...B_{n}$, and $A_{n}\rightarrow A$,
$B_{n}\rightarrow B$ as $n\rightarrow\infty$. ∎

##  8\. Conclusion

The problem in using transformation technique lies in guessing methods for
transformation. For example, transformations were not used since nineteenth
century to define multiplicative modulus function from additive modulus
function. One can guess that a transformation can be applied to derive
multiplicative modulus function from additive modulus function, only after
introducing the concept of multiplicative modulus function. So, separate
techniques should be developed for infinite products. Theory of infinite
products is also applicable like theory of infinite series. This is the need
for development of theory of infinite products. Corollary 7.4 is simple to
write examples. But it is not sufficient. So, Theorem 5.9 should be improved
in all possible ways, but subject to simple conditions.

Declaration: There is no conflict of interest.

### Acknowledgement

Dr. C. Ganesa Moorthy (Professor, Department of Mathematics, Alagappa
University, Karaikudi- 630003, INDIA) gratefully acknowledges the joint
financial support of RUSA-Phase 2.0 grant sanctioned vide letter No.F
24-51/2014-U, Policy (TN Multi-Gen), Dept. of Edn. Govt. of India, Dt.
09.10.2018, UGC-SAP (DRS-I) vide letter No.F.510/8/DRS-I/2016 (SAP-I) Dt.
23.08.2016 and DST (FIST - level I) 657876570 vide letter
No.SR/FIST/MS-I/2018-17 Dt. 20.12.2018.

## References

  * [1] T. M. Apostol, Mathematical Analysis, 2nd ed., Addison-Wesley, Reading, 1974.
  * [2] J. L. Kelly, General Topology, Van Nostrand, New York, 1955.
  * [3] S. Loganathan and C. G. Moorthy, A net convergence for Schauder double bases, Asian-European J. Math., 9(1) (2016), 1-33.