# Spectral properties of balanced trees and dendrimers

Ivan Damnjanović
University of Niš, Faculty of Electronic Engineering
Aleksandra Medvedeva 14, 18115 Niš, Serbia
<EMAIL_ADDRESS>
The corresponding author. Slobodan Filipovski
University of Primorska, Faculty of Mathematics,
Natural Sciences and Information Technologies
Glagoljaška 8, 6000 Koper, Slovenia
<EMAIL_ADDRESS>
The second author is supported by the Slovenian Research Agency through the
grants P1-0285, J1-1695, J1-9108 and J1-9110, as well as the bilateral project
337-00-21/2020-09/28 between the Slovenian Research Agency and the Ministry of
Education, Science and Technological Development of the Republic of Serbia.
Dragan Stevanović
Mathematical Institute of the Serbian Academy of Sciences and Arts
Kneza Mihaila 36, 11000 Belgrade, Serbia
<EMAIL_ADDRESS>The third author is supported by the Serbian
Academy of Sciences and Arts through the grant F-159 and by the Ministry of
Education, Science and Technological Development of the Republic of Serbia
through the Mathematical Institute of SASA, as well as the bilateral project
337-00-21/2020-09/28 between the Slovenian Research Agency and the Ministry of
Education, Science and Technological Development of the Republic of Serbia.

###### Abstract

We investigate the spectral properties of balanced trees and dendrimers, with
a view toward unifying and improving the existing results. Here we find a
semi-factorized formula for their characteristic polynomials. Afterwards, we
determine their spectra via the aforementioned factors. In the end, we analyze
the behavior of the energy of dendrimers and compute lower and upper bound
approximations for it.

Mathematics Subject Classification: 05C50, 05C05, 05C92, 40A05.
Keywords: Balanced tree, Dendrimer, Characteristic polynomial, Spectrum, Graph
energy.

##  1 Introduction

Let $G$ be a simple graph with $n$ vertices and the adjacency matrix $A(G)$.
Let $P_{x}(G)=\det(xI-A(G))$ denote the characteristic polynomial of $A(G)$,
and let $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ be the eigenvalues of
$A(G)$. The energy of $G$ is defined as
$E(G)=\sum_{j=1}^{n}\lvert\lambda_{j}\rvert$, as introduced by Gutman in [1].
Let $\sigma^{*}(G)$ represent the set of all the distinct eigenvalues of
$A(G)$.

A balanced tree is an unweighted rooted tree such that all the vertices from
the same level have an equal degree, while a dendrimer is defined as a
balanced tree whose internal vertices all have the same degree. We will use
$d(l,k)$ to denote the dendrimer which contains $l+1$ levels enumerated from
$0$ to $l$ with each internal vertex having a degree of $k$. More about the
various applications of dendrimers can be found, for example, in [2, 3].

In this paper, we investigate the spectral properties of balanced trees and
dendrimers. The two main results we obtain deal with the computation of
$E(d(l,k))$ and are given as follows:

######  Theorem 1.

For any fixed value of $l\geq 1$, we have

$E(d(l,k))\sim 2(k-1)^{l-1/2}\qquad\mbox{as $\ k\to\infty$}\,.$

Also, for any fixed value of $k\geq 3$, we have

$E(d(l,k))\sim\mu_{k}(k-1)^{l-1/2}\qquad\mbox{as $\ l\to\infty$}\,.$

where $\mu_{k}$ is the positive real number which represents the sum of the
convergent positive series

$\sum_{j=0}^{\infty}f_{j}(k-1)^{-j}$

defined by

$f_{j}=\begin{cases}2\csc\left(\dfrac{\pi}{2j+6}\right)-2\csc\left(\dfrac{\pi}{2j+2}\right),&2\mid
j\,,\\\
2\cot\left(\dfrac{\pi}{2j+6}\right)-2\cot\left(\dfrac{\pi}{2j+2}\right),&2\nmid
j\,.\end{cases}$

######  Theorem 2.

For a given dendrimer $d(l,k)$, where $k\geq 3$ and $l\geq 2$, we have

$\displaystyle E(d(l,k))$
$\displaystyle<(k-1)^{l-1/2}\left(2+\dfrac{0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{k-1}\right),$
$\displaystyle E(d(l,k))$
$\displaystyle>(k-1)^{l-1/2}\left(2+\dfrac{2\sqrt{2}}{k-1}\right).$

The remainder of this paper is structured as follows. In Section 2 we prove a
result which yields the characteristic polynomials and spectra of balanced
trees, improving the previous result of Rojo and Soto [4]. We further focus on
computing the characteristic polynomials and spectra of dendrimers in Sections
3–4. The results in these sections improve the previous results of Stevanović
[5] and Bokhary and Tabassum [6]. In Section 5 we deal with the computation of
the energy of dendrimers and use the results from Sections 3–4 in order to
prove the two main theorems, namely Theorem 1 and Theorem 2.

##  2 Characteristic polynomials and eigenvalues of balanced trees

Let $T$ be a balanced tree. First, we can uniquely represent $T$ up to
isomorphism by assigning to it a characteristic tuple of positive integers
$C_{T}=(c_{1},c_{2},\ldots,c_{l})$. The number of elements in this tuple
dictates how many internal levels $T$ has. We will enumerate the tree levels
from $0$ to $l$, where the levels $0,1,\ldots,l-1$ are internal, while the
level $l$ contains the leaves. In the tuple $C_{T}$, the element $c_{j}$
describes the number of children of each vertex in level $j-1$. In other
words, $c_{1}$ represents the degree of the root of $T$, while $c_{j}+1$
determines the degree of each vertex in level $j-1$, for all $2\leq j\leq l$.

Furthermore, we are going to use $n_{j}$ to denote the total number of
vertices in level $j$, for $0\leq j\leq l$. Here we obviously have that
$n_{0}=1$ and $n_{j}=n_{j-1}c_{j}$ for all $1\leq j\leq l$. Also, let
$n_{T}=\sum_{j=0}^{l}n_{j}$ so that $n_{T}$ represents the total number of
vertices in $T$. For convenience, we will define $n_{-1}=0$.

The following theorem describes the characteristic polynomials of arbitrary
balanced trees.

######  Theorem 3.

Let $T$ be a balanced tree such that its characteristic tuple equals
$C_{T}=(c_{1},c_{2},\ldots,c_{l})$. If we define a sequence of polynomials
$Q_{0}(x),Q_{1}(x),\ldots,\linebreak Q_{l+1}(x)$ by

$\displaystyle Q_{0}(x)$ $\displaystyle=1\,,$ $\displaystyle Q_{1}(x)$
$\displaystyle=x\,,$ $\displaystyle Q_{j+2}(x)$
$\displaystyle=x\,Q_{j+1}(x)-c_{l-j}\,Q_{j}(x)\qquad\mbox{for all $\,0\leq
j\leq l-1$}\,,$

then

$P_{x}(T)=\prod_{j=1}^{l+1}Q_{j}(x)^{n_{l+1-j}-n_{l-j}}\,.$

Before we prove the theorem, we will make certain preliminary definitions
which will aid us in creating a more concise proof. For two given positive
integers $\alpha$ and $\beta$ such that $\beta\mid\alpha$, we will use
$B_{\alpha,\beta}=(b_{ij})_{\alpha\times\beta}\in\mathbb{R}^{\alpha\times\beta}$
to denote the binary matrix whose rows and columns are enumerated from $0$ to
$\alpha-1$ and $0$ to $\beta-1$ respectively, such that $b_{ij}=1$ if and only
if $\lfloor\frac{i\beta}{\alpha}\rfloor=j$. For example, we have

$\displaystyle B_{8,2}=\begin{bmatrix}1&0\\\ 1&0\\\ 1&0\\\ 1&0\\\ 0&1\\\
0&1\\\ 0&1\\\ 0&1\end{bmatrix},$ $\displaystyle
B_{9,3}=\begin{bmatrix}1&0&0\\\ 1&0&0\\\ 1&0&0\\\ 0&1&0\\\ 0&1&0\\\ 0&1&0\\\
0&0&1\\\ 0&0&1\\\ 0&0&1\end{bmatrix},$ $\displaystyle
B_{10,5}=\begin{bmatrix}1&0&0&0&0\\\ 1&0&0&0&0\\\ 0&1&0&0&0\\\ 0&1&0&0&0\\\
0&0&1&0&0\\\ 0&0&1&0&0\\\ 0&0&0&1&0\\\ 0&0&0&1&0\\\ 0&0&0&0&1\\\
0&0&0&0&1\end{bmatrix}.$

We can concisely express the adjacency matrix of $T$ with the help of these
matrices:

$A(T)=\begin{bmatrix}\mathbf{O}&B_{n_{l},n_{l-1}}&\mathbf{O}&\cdots&\mathbf{O}&\mathbf{O}\\\
B_{n_{l},n_{l-1}}^{T}&\mathbf{O}&B_{n_{l-1},n_{l-2}}&\cdots&\mathbf{O}&\mathbf{O}\\\
\mathbf{O}&B_{n_{l-1},n_{l-2}}^{T}&\mathbf{O}&\cdots&\mathbf{O}&\mathbf{O}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&\mathbf{O}&B_{n_{1},n_{0}}\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&B_{n_{1},n_{0}}^{T}&\mathbf{O}\end{bmatrix}\,.$

Together with $P_{x}(T)=\det(x\,\mathbf{I}_{n}-A(T))$, this leads us to

$P_{x}(T)=\begin{vmatrix}x\,\mathbf{I}_{n_{l}}&-B_{n_{l},n_{l-1}}&\mathbf{O}&\cdots&\mathbf{O}&\mathbf{O}\\\
-B_{n_{l},n_{l-1}}^{T}&x\,\mathbf{I}_{n_{l-1}}&-B_{n_{l-1},n_{l-2}}&\cdots&\mathbf{O}&\mathbf{O}\\\
\mathbf{O}&-B_{n_{l-1},n_{l-2}}^{T}&x\,\mathbf{I}_{n_{l-2}}&\cdots&\mathbf{O}&\mathbf{O}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&x\,\mathbf{I}_{n_{1}}&-B_{n_{1},n_{0}}\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&-B_{n_{1},n_{0}}^{T}&x\,\mathbf{I}_{n_{0}}\end{vmatrix}.$
(1)

We will compute $P_{x}(T)$ from Eq. (1).

Proof of Theorem 3. By multiplying the $(j+1)$-th block row in Eq. (1) with
$Q_{j}(x)$, for all $1\leq j\leq l$, we obtain

$\displaystyle P_{x}(T)\prod_{j=1}^{l}Q_{j}(x)^{n_{l-j}}=$
$\displaystyle=\begin{vmatrix}x\,Q_{0}(x)\mathbf{I}_{n_{l}}&-Q_{0}(x)B_{n_{l},n_{l-1}}&\mathbf{O}&\cdots&\mathbf{O}\\\
-Q_{1}(x)B_{n_{l},n_{l-1}}^{T}&x\,W_{1}(x)\mathbf{I}_{n_{l-1}}&-Q_{1}(x)B_{n_{l-1},n_{l-2}}&\cdots&\mathbf{O}\\\
\mathbf{O}&-Q_{2}(x)B_{n_{l-1},n_{l-2}}^{T}&x\,Q_{2}(x)\mathbf{I}_{n_{l-2}}&\cdots&\mathbf{O}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&x\,Q_{l}(x)\mathbf{I}_{n_{0}}\end{vmatrix}\,.$

The determinant on the right-hand side can easily be calculated by applying
the Gaussian elimination on the given block matrix. If we multiply the first
block row by $B_{n_{l},n_{l-1}}^{T}$ to the left and then add the obtained
result to the second row, we get

$\displaystyle P_{x}(T)\prod_{j=1}^{l}Q_{j}(x)^{n_{l-j}}=$
$\displaystyle=\begin{vmatrix}x\,Q_{0}(x)\mathbf{I}_{n_{l}}&-Q_{0}(x)B_{n_{l},n_{l-1}}&\mathbf{O}&\cdots&\mathbf{O}\\\
\mathbf{O}&Q_{2}(x)\mathbf{I}_{n_{l-1}}&-Q_{1}(x)B_{n_{l-1},n_{l-2}}&\cdots&\mathbf{O}\\\
\mathbf{O}&-Q_{2}(x)B_{n_{l-1},n_{l-2}}^{T}&x\,Q_{2}(x)\mathbf{I}_{n_{l-2}}&\cdots&\mathbf{O}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&x\,Q_{l}(x)\mathbf{I}_{n_{0}}\end{vmatrix}\,.$

We can then multiply the second block row by $B_{n_{l-1},n_{l-2}}^{T}$ to the
left and add the result to the third row, and so on and so forth, until we
reach

$\displaystyle P_{x}(T)\prod_{j=1}^{l}Q_{j}(x)^{n_{l-j}}=$
$\displaystyle=\begin{vmatrix}Q_{1}(x)\mathbf{I}_{n_{l}}&-Q_{0}(x)B_{n_{l},n_{l-1}}&\mathbf{O}&\cdots&\mathbf{O}\\\
\mathbf{O}&Q_{2}(x)\mathbf{I}_{n_{l-1}}&-Q_{1}(x)B_{n_{l-1},n_{l-2}}&\cdots&\mathbf{O}\\\
\mathbf{O}&\mathbf{O}&Q_{3}(x)\mathbf{I}_{n_{l-2}}&\cdots&\mathbf{O}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\\\
\mathbf{O}&\mathbf{O}&\mathbf{O}&\cdots&Q_{l+1}(x)\mathbf{I}_{n_{0}}\end{vmatrix}\,,$

which finally gives us

$P_{x}(T)\prod_{j=1}^{l}Q_{j}(x)^{n_{l-j}}=\prod_{j=1}^{l+1}Q_{j}(x)^{n_{l+1-j}}\,.$

By taking into consideration that $n_{-1}=0$, we get that for any complex
number $z$ which is not a root of any of the polynomials
$Q_{1}(x),Q_{2}(x),\ldots,Q_{l}(x)$, the following equality must hold:

$P_{z}(T)=\prod_{j=1}^{l+1}Q_{j}(z)^{n_{l+1-j}-n_{l-j}}\,.$ (2)

Since each polynomial $Q_{j}(x)$ is of degree $j$, for all $1\leq j\leq l$, we
conclude that there are infinitely many complex numbers which are not a root
of any of these polynomials. From Eq. (2), we see that the polynomials
$P_{x}(T)$ and $\prod_{j=1}^{l+1}Q_{j}(x)^{n_{l+1-j}-n_{l-j}}$ must be equal
in infinitely many points, which implies that these two polynomials are
identical. ∎

By Theorem 3, $P_{x}(T)$ is a product of those $Q_{j}(x)$ for which
$n_{l+1-j}-n_{l-j}$ is positive. This implies the following result.

######  Theorem 4.

Let $\Phi$ be the set of positive integers $1\leq j\leq l+1$ such that
$n_{l+1-j}>n_{l-j}$. Then we have

$\sigma^{*}(T)=\bigcup_{j\in\Phi}\,\\{x\colon Q_{j}(x)=0\\}.$

######  Remark 5.

The condition $n_{l+1-j}>n_{l-j}$ is always satisfied for $j=l+1$. For $j\leq
l$, it becomes equivalent to $c_{l+1-j}\neq 1$.

The use of Theorem 3 is exemplified in the following section on the case of
dendrimers.

##  3 Characteristic polynomials of dendrimers

If we view a dendrimer $d(l,k)$ as a balanced tree, we see that its
characteristic tuple equals $(k,\underbrace{k-1,k-1,\ldots,k-1}_{\text{ $l-1$
times}})$. We can now apply Theorem 3 in order to compute the characteristic
polynomial of $d(l,k)$.

######  Theorem 6.

Let $W_{l,0}(x,k),W_{l,1}(x,k),\ldots,W_{l,l+1}(x,k)$ be a sequence of
polynomials defined in the following manner

$\displaystyle\begin{split}W_{l,0}(x,k)&=1\,,\\\ W_{l,1}(x,k)&=x\,,\\\
W_{l,j}(x,k)&=x\,W_{l,j-1}(x,k)-(k-1)\,W_{l,j-2}(x,k)\qquad\mbox{for all
$\,2\leq j\leq l$}\,,\\\
W_{l,l+1}(x,k)&=x\,W_{l,l}(x,k)-k\,W_{l,l-1}(x,k)\,.\end{split}$ (3)

Then, for all $k\geq 2$ and $l\geq 1$,

$P_{x}(d(l,k))=W_{l,l+1}(x,k)W_{l,l}(x,k)^{k-1}\prod_{j=1}^{l-1}W_{l,j}(x,k)^{k(k-2)(k-1)^{l-1-j}}\,.$
(4)

###### Proof.

Given the fact that $c_{1}=k$ and $c_{j}=k-1$ for all $2\leq j\leq l$, Theorem
3 gives us

$P_{x}(T)=\prod_{j=1}^{l+1}W_{l,j}(x,k)^{n_{l+1-j}-n_{l-j}}\,.$

Since $n_{-1}=0,n_{0}=1$ and $n_{j}=k(k-1)^{j-1}$ for all $1\leq j\leq l$, we
have $n_{0}-n_{-1}=1,n_{1}-n_{0}=k-1$ and
$n_{j+1}-n_{j}=k(k-1)^{j}-k(k-1)^{j-1}$ for all $1\leq j\leq l-1$. This leads
us to

$\displaystyle P_{x}(T)$
$\displaystyle=W_{l,l+1}(x,k)^{n_{0}-n_{-1}}W_{l,l}(x,k)^{n_{1}-n_{0}}\prod_{j=1}^{l-1}W_{l,j}(x,k)^{n_{l+1-j}-n_{l-j}}$
$\displaystyle=W_{l,l+1}(x,k)W_{l,l}(x,k)^{k-1}\prod_{j=1}^{l-1}W_{l,j}(x,k)^{k(k-1)^{l-j}-k(k-1)^{l-j-1}}$
$\displaystyle=W_{l,l+1}(x,k)W_{l,l}(x,k)^{k-1}\prod_{j=1}^{l-1}W_{l,j}(x,k)^{k(k-1)^{l-j-1}(k-2)}\,.$

∎

It is relatively easy to use Theorem 6 to obtain the characteristic polynomial
of the dendrimer $d(l,k)$, where $k\geq 2$ can be treated as an integer
parameter, while $l\geq 1$ is some fixed and preferably small, positive
integer. For example, if we put $l=1$, we will get a sequence of polynomials

$\displaystyle W_{1,0}(x,k)$ $\displaystyle=1\,,$ $\displaystyle W_{1,1}(x,k)$
$\displaystyle=x\,,$ $\displaystyle W_{1,2}(x,k)$ $\displaystyle=x^{2}-k\,,$

which quickly gives us the characteristic polynomial

$P_{x}(d(1,k))=(x^{2}-k)x^{k-1}\,.$ (5)

Similarly, by setting $l=2$ we obtain the sequence of polynomials

$\displaystyle W_{2,0}(x,k)$ $\displaystyle=1\,,$ $\displaystyle W_{2,1}(x,k)$
$\displaystyle=x\,,$ $\displaystyle W_{2,2}(x,k)$
$\displaystyle=x^{2}-(k-1)\,,$ $\displaystyle W_{2,3}(x,k)$
$\displaystyle=x^{3}-(2k-1)x\,,$

which leads to the characteristic polynomial

$\displaystyle P_{x}(d(2,k))$
$\displaystyle=[x^{3}-(2k-1)x][x^{2}-(k-1)]^{k-1}x^{k(k-2)}$
$\displaystyle=x^{k^{2}-2k+1}(x^{2}-k+1)^{k-1}(x^{2}-2k+1)$
$\displaystyle=x^{(k-1)^{2}}(x^{2}-k+1)^{k-1}(x^{2}-2k+1)\,.$ (6)

Further for $l=3$ we get the sequence of polynomials

$\displaystyle W_{3,0}(x,k)$ $\displaystyle=1\,,$ $\displaystyle W_{3,1}(x,k)$
$\displaystyle=x\,,$ $\displaystyle W_{3,2}(x,k)$
$\displaystyle=x^{2}-(k-1)\,,$ $\displaystyle W_{3,3}(x,k)$
$\displaystyle=x^{3}-(2k-2)x\,,$ $\displaystyle W_{3,4}(x,k)$
$\displaystyle=x^{4}-(3k-2)x^{2}+k(k-1)\,,$

which yields the characteristic polynomial

$\displaystyle P_{x}(d(3,k))={}$
$\displaystyle[x^{4}-(3k-2)x^{2}+k(k-1)][x^{3}-(2k-2)x]^{k-1}$
$\displaystyle[x^{2}-(k-1)]^{k(k-2)}x^{k(k-2)(k-1)}$ $\displaystyle={}$
$\displaystyle[x^{4}-(3k-2)x^{2}+k(k-1)](x^{2}-2k+2)^{k-1}$
$\displaystyle(x^{2}-k+1)^{k(k-2)}x^{k(k-2)(k-1)+(k-1)}$ $\displaystyle={}$
$\displaystyle[x^{4}-(3k-2)x^{2}+k(k-1)](x^{2}-2k+2)^{k-1}$
$\displaystyle(x^{2}-k+1)^{k(k-2)}x^{(k-1)^{3}}.$

######  Remark 7.

Note that [6, Theorem 3.5] gives a slightly inaccurate expression for
$P_{x}(d(3,k))$, in which the correct factor $x^{4}-(3k-2)x^{2}+k(k-1)$ above
is replaced by an incorrect factor $x^{4}-2(k+1)x^{2}+4(k-1)$.

##  4 Eigenvalues of dendrimers

Theorem 6 can be used to determine the eigenvalues of the given dendrimer
$d(l,k)$. If $k>2$, then a real number belongs to $\sigma^{*}(d(l,k))$ if and
only if it represents a root of some polynomial from the sequence
$W_{l,1}(x,k),W_{l,2}(x,k),\ldots,W_{l,l+1}(x,k)$, given the fact that
$c_{j}>1$ for all $1\leq j\leq l$. A special case occurs when $k=2$, since
this would lead to $c_{2}=c_{3}=\cdots=c_{l}=1$. In this situation, we would
have that $\sigma^{*}(d(l,k))$ is composed solely of real numbers which are a
root of $W_{l,l+1}(x,k)$ or $W_{l,l}(x,k)$.

######  Theorem 8.

For any $k\geq 3$ and $l\geq 1$, we have

$\sigma^{*}(d(l,k))=\bigcup_{j=1}^{l+1}\,\\{x\colon W_{l,j}(x,k)=0\\}.$

If $k=2$ and $l\geq 1$, then we get

$\sigma^{*}(d(l,2))=\\{x\colon W_{l,l+1}(x,k)=0\vee W_{l,l}(x,k)=0\\}.$

A natural question is whether these results can be made explicit. The answer
to this question is affirmative, as previously shown by one of the present
authors in [5]. Let $E_{0}(x,a),E_{1}(x,a),E_{2}(x,a),\ldots$ be a sequence of
polynomials defined via the recurrence relation

$\displaystyle\begin{split}E_{0}(x,a)&=1\,,\\\ E_{1}(x,a)&=x\,,\\\
E_{j}(x,a)&=x\,E_{j-1}(x,a)-a\,E_{j-2}(x,a)\qquad\mbox{for all $\,j\geq
2$}\,.\\\ \end{split}$ (7)

We shall call these polynomials the Dickson polynomials of the second kind, as
done so in [7, pp. 9–10]. By comparing Eqs. (7) and (3), we see that the
polynomials $W_{l,0}(x,k),W_{l,1}(x,k),\ldots,W_{l,l}(x,k)$ actually represent
the Dickson polynomials of the second kind
$E_{0}(x,k-1),E_{1}(x,k-1),\ldots,E_{l}(x,k-1)$, respectively. This means that
the polynomial $W_{l,j}(x,k)$ must have $j$ distinct simple roots
$2\sqrt{k-1}\cos\left(\frac{1}{j+1}\pi\right),2\sqrt{k-1}\cos\left(\frac{2}{j+1}\pi\right),\ldots,2\sqrt{k-1}\cos\left(\frac{j}{j+1}\pi\right)$,
for each $0\leq j\leq l$ (see, for example, [7, pp. 9–10]).

Note that the roots of $W_{l,l+1}(x,k)$ cannot be found as easily, given the
fact that this polynomial is not a Dickson polynomial of the second kind,
unlike all of its predecessors. This irregularity occurs due to the fact that
$W_{l,l+1}(x,k)$ is defined via a recurrence relation which is slightly
different from all of the previous ones.

By taking these facts into consideration, we reach the following theorem.

######  Theorem 9.

For any $k\geq 3$ and $l\geq 1$, we have

$\displaystyle\sigma^{*}(d(l,k))={}$
$\displaystyle\left\\{2\sqrt{k-1}\cos\left(\frac{h}{j+1}\pi\right)\colon 1\leq
h\leq j\leq l\right\\}$ $\displaystyle\cup\,\\{x\in\mathbb{R}\colon
W_{l,l+1}(x,k)=0\\}\,.$

On the other hand, if $k=2$ and $l\geq 1$, then

$\sigma^{*}(d(l,2))=\left\\{2\cos\left(\frac{h}{2l+2}\pi\right)\colon 1\leq
h\leq 2l+1\right\\}\,.$

###### Proof.

The dendrimer $d(l,2)$ is just a path graph composed of $2l+1$ vertices. It is
known (see, for example, [8, pp. 18]) that the spectrum of this graph is
composed of the real numbers
$2\cos\left(\frac{1}{2l+2}\pi\right),2\cos\left(\frac{2}{2l+2}\pi\right),\ldots,2\cos\left(\frac{2l+1}{2l+2}\pi\right)$.

For $k\geq 3$, from $W_{l,j}(x,k)\equiv E_{j}(x,k-1)$ for $0\leq j\leq l$, we
have

$\bigcup_{j=1}^{l}\,\\{x\in\mathbb{R}\colon
W_{j}(x)=0\\}=\left\\{2\sqrt{k-1}\cos\left(\frac{h}{j+1}\pi\right)\colon 1\leq
h\leq j\leq l\right\\}.$ (8)

∎

Despite the fact that the roots of $W_{l,l+1}(x,k)$ cannot explicitly be
found, a relatively good approximation can be made. Moreover, $W_{l,l+1}(x,k)$
represents a Geronimus polynomial of degree $l+1$ (see, for example, [9]).
This fact can be shown if we notice that

$\displaystyle W_{l,l+1}(x,k)$
$\displaystyle=x\,W_{l,l}(x,k)-k\,W_{l,l-1}(x,k)$
$\displaystyle=x\,E_{l}(x,k-1)-(k-1)\,E_{l-1}(x,k-1)-E_{l-1}(x,k-1)$
$\displaystyle=E_{l+1}(x,k-1)-E_{l-1}(x,k-1)\,,$

for each $l\geq 1$.

For a fixed parameter $a\in\mathbb{R}$, we will use $G_{j}(x,a)$ to denote the
sequence of Geronimus polynomials defined via

$\displaystyle G_{0}(x,a)$ $\displaystyle=1\,,$ $\displaystyle G_{1}(x,a)$
$\displaystyle=x\,,$ $\displaystyle G_{2}(x,a)$ $\displaystyle=x^{2}-a\,,$
$\displaystyle G_{j}(x,a)$
$\displaystyle=xG_{j-1}(x,a)-(a-1)G_{j-2}(x,c)\qquad\mbox{for all $\,j\geq
3$}\,.$

It is easily proven via mathematical induction that
$G_{j}(x,a)=E_{j}(x,a-1)-E_{j-2}(x,a-1)$ for each $j\geq 2$, which ultimately
shows that $W_{l,l+1}(x,k)\equiv G_{l+1}(x,k)$. This observation leads to the
following lemma.

######  Lemma 10.

For $k\geq 3$, the polynomial $W_{l,l+1}(x,k)$ has $l+1$ simple real roots
determined by the set

$\\{\pm\alpha_{1},\pm\alpha_{2},\ldots,\pm\alpha_{\lfloor(l+1)/2\rfloor}\\}$

if $l+1$ is even, or

$\\{\pm\alpha_{1},\pm\alpha_{2},\ldots,\pm\alpha_{\lfloor(l+1)/2\rfloor}\\}\cup\\{0\\}$

if $l+1$ is odd, where the numbers
$\alpha_{1},\alpha_{2},\ldots,\alpha_{\lfloor(l+1)/2\rfloor}$ form a strictly
decreasing positive sequence which satisfies

$2\sqrt{k-1}\cos\left(\dfrac{j-0.5}{l+2}\pi\right)>\alpha_{j}>2\sqrt{k-1}\cos\left(\dfrac{j+0.5}{l+2}\pi\right)$

for each $j\in\\{1,2,3,\ldots,\lfloor(l+1)/2\rfloor\\}$.

###### Proof.

Given the fact that $W_{l,l+1}(x,k)\equiv G_{l+1}(x,k)$, we know from [9] that
for $k\geq 3$ the polynomial $W_{l,l+1}(x,k)$ must have $l+1$ distinct real
roots $\alpha_{1},\alpha_{2},\ldots,\alpha_{l+1}$ such that

$2\sqrt{k-1}\cos\left(\dfrac{j-0.5}{l+2}\pi\right)>\alpha_{j}>2\sqrt{k-1}\cos\left(\dfrac{j+0.5}{l+2}\pi\right)$
(9)

for each $j\in\\{1,2,3,\ldots,l+1\\}$. Since the cosine is a strictly
decreasing function on $[0,\pi]$, Eq. (9) immediately shows that

$\alpha_{1}>\alpha_{2}>\alpha_{3}>\cdots>\alpha_{l}>\alpha_{l+1}\,.$

Also, due to the fact that the cosine takes positive values on $[0,\pi/2)$ and
negative values on $(\pi/2,\pi]$, we obtain

$\alpha_{1}>\alpha_{2}>\cdots>\alpha_{(l+1)/2}>0>\alpha_{(l+1)/2+1}>\cdots>\alpha_{l}>\alpha_{l+1}$
(10)

if $l+1$ is even, and

$\alpha_{1}>\alpha_{2}>\cdots>\alpha_{l/2}>0>\alpha_{l/2+2}>\cdots>\alpha_{l}>\alpha_{l+1}$
(11)

if $l+1$ is odd.

It is trivial to prove that $G_{j}(x,k)$ is an even polynomial for each even
$j$ and an odd polynomial for each odd $j$. If $l+1$ is even, this observation
and Eq. (10) together give that $\alpha_{l+2-j}=-\alpha_{j}$ for each $1\leq
j\leq(l+1)/2$. This implies that the roots of $W_{l,l+1}(x,k)$ are described
via set
$\\{\pm\alpha_{1},\pm\alpha_{2},\ldots,\pm\alpha_{\lfloor(l+1)/2\rfloor}\\}$,
where $\alpha_{1}>\alpha_{2}>\cdots>\alpha_{\lfloor(l+1)/2\rfloor}>0$.

If $l+1$ is odd, then $0$ must be a root of $W_{l,l+1}(x,k)$, which implies
$\alpha_{l/2+1}=0$. On the other hand, Eq. (11) gives
$\alpha_{l+2-j}=-\alpha_{j}$ for each $1\leq j\leq l/2$. We conclude that in
this case the roots of $W_{l,l+1}(x,k)$ can be described by the set
$\\{\pm\alpha_{1},\pm\alpha_{2},\ldots,\pm\alpha_{\lfloor(l+1)/2\rfloor}\\}\cup\\{0\\}$,
where $\alpha_{1}>\alpha_{2}>\cdots>\alpha_{\lfloor(l+1)/2\rfloor}>0$. ∎

Lemma 10 shall prove to be quite useful while approximating the energy of
dendrimers in the following section.

##  5 Energy of dendrimers

Here we discuss the computation of the energy of dendrimers. Due to the fact
that the roots of the Geronimus polynomials cannot explicitly be found, we are
unable to obtain an exact expression for the energy of a dendrimer $d(l,k)$
whenever $k\geq 3$. Instead, we compute a reasonable approximation of
$E(d(l,k))$ that covers all of the dendrimers whenever $k\geq 3$ and $l\geq
1$, together with the exact value of $E(d(l,2))$ for each $l\geq 1$.
Afterwards, we inspect the asymptotic behavior of $E(d(l,k))$ when analyzed as
a function of two variables $l\in\mathbb{N}$ and
$k\in\mathbb{N}\setminus\\{1,2\\}$. In the end, we give another approximative
formula of $E(d(l,k))$ when $k\geq 3$, which reflects the asymptotic behavior
of the function. To begin with, we state the following theorem:

######  Theorem 11.

For any $k\geq 3$, the energy of the dendrimer $d(l,k)$ can be approximated in
the following manner:

$\displaystyle E(d(l,k))$
$\displaystyle<\sum_{j=0}^{l-1}f_{j}(k-1)^{(l-1/2)-j}+2(k-1)^{1/2}\,,$ (12)
$\displaystyle E(d(l,k))$
$\displaystyle>\sum_{j=0}^{l-1}f_{j}(k-1)^{(l-1/2)-j}-2.4(k-1)^{1/2}\,,$ (13)

where

$f_{j}=\begin{cases}2\csc\left(\dfrac{\pi}{2j+6}\right)-2\csc\left(\dfrac{\pi}{2j+2}\right),&2\mid
j\,,\\\
2\cot\left(\dfrac{\pi}{2j+6}\right)-2\cot\left(\dfrac{\pi}{2j+2}\right),&2\nmid
j\,.\end{cases}$

Also, the energy of the dendrimer $d(l,2)$ can be computed via:

$E(d(l,2))=2\left(\cot\left(\dfrac{\pi}{4l+4}\right)-1\right).$ (14)

In order to make the proof of Theorem 11 easier to follow, we shall introduce
and prove two auxiliary lemmas. First of all, let $\Psi(W_{l,j}(x,k))$ denote
the sum of absolute values of all the roots of the polynomial $W_{l,j}(x,k)$.

######  Lemma 12.

For each $0\leq j\leq l$, we have

$\Psi(W_{l,j}(x,k))=2\sqrt{k-1}\left(\cot\left(\dfrac{\pi}{2j+2}\right)-1\right)$

if $j$ is odd, and

$\Psi(W_{l,j}(x,k))=2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2j+2}\right)-1\right)$

if $j$ is even.

###### Proof.

The equality is directly proven for $j=0$. The polynomial $W_{l,0}(x,k)=1$ is
of degree $0$ and has no roots, hence $\Psi(W_{l,0}(x,k))=0$, while the
according right-hand side
$2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2}\right)-1\right)$ also obviously
equals $0$.

Suppose $1\leq j\leq l$. We know that the roots of the polynomial
$W_{l,j}(x,k)$ must be
$2\sqrt{k-1}\cos\left(\frac{1}{j+1}\pi\right),2\sqrt{k-1}\cos\left(\frac{2}{j+1}\pi\right),\ldots,2\sqrt{k-1}\cos\left(\frac{j}{j+1}\pi\right)$.
Hence

$\displaystyle\Psi(W_{l,j}(x,k))$
$\displaystyle=\sum_{h=1}^{j}\left|2\sqrt{k-1}\cos\left(\frac{h}{j+1}\pi\right)\right|$
$\displaystyle=2\sqrt{k-1}\sum_{h=1}^{j}\left|\cos\left(\frac{h}{j+1}\pi\right)\right|.$

Since $\cos\left(\frac{h}{j+1}\pi\right)>0$ for $1\leq h<\frac{j+1}{2}$ and
$\cos\left(\frac{h}{j+1}\pi\right)=-\cos\left(\frac{j+1-h}{j+1}\pi\right)$ for
all $\frac{j+1}{2}<h\leq j$, we can rewrite the last expression as

$\Psi(W_{l,j}(x,k))=4\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{j}{2}\rfloor}\cos\left(\frac{h}{j+1}\pi\right).$

Let us denote $\zeta=e^{\frac{i\pi}{j+1}}$. It is convenient to replace
$\cos\left(\frac{h}{j+1}\pi\right)$ with $\frac{\zeta^{h}+\zeta^{-h}}{2}$.
This gives us:

$\displaystyle\Psi(W_{l,j}(x,k))$
$\displaystyle=4\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{j}{2}\rfloor}\frac{\zeta^{h}+\zeta^{-h}}{2}$
$\displaystyle=2\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{j}{2}\rfloor}\left(\zeta^{h}+\zeta^{-h}\right)$
$\displaystyle=2\sqrt{k-1}\left(\left(\sum_{h=-\lfloor\frac{j}{2}\rfloor}^{\lfloor\frac{j}{2}\rfloor}\zeta^{h}\right)-1\right).$

Since $\zeta\neq 1$, we can use the standard formula for summing a geometric
progression in order to get

$\displaystyle\Psi(W_{l,j}(x,k))$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\sum_{h=0}^{2\lfloor\frac{j}{2}\rfloor}\zeta^{h}}{\zeta^{\lfloor\frac{j}{2}\rfloor}}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\zeta^{2\lfloor\frac{j}{2}\rfloor+1}-1}{\zeta^{\lfloor\frac{j}{2}\rfloor}(\zeta-1)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\zeta^{\lfloor\frac{j}{2}\rfloor+1}-\zeta^{-\lfloor\frac{j}{2}\rfloor}}{\zeta-1}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\left(\zeta^{\lfloor\frac{j}{2}\rfloor+1}-\zeta^{-\lfloor\frac{j}{2}\rfloor}\right)\left(\frac{1}{\zeta}-1\right)}{(\zeta-1)\left(\frac{1}{\zeta}-1\right)}-1\right).$

By taking into consideration that

$\displaystyle\left(\zeta^{\lfloor\frac{j}{2}\rfloor+1}-\zeta^{-\lfloor\frac{j}{2}\rfloor}\right)\left(\frac{1}{\zeta}-1\right)$
$\displaystyle=\zeta^{\lfloor\frac{j}{2}\rfloor}+\zeta^{-\lfloor\frac{j}{2}\rfloor}-\zeta^{\lfloor\frac{j}{2}\rfloor+1}-\zeta^{-\lfloor\frac{j}{2}\rfloor-1}$
$\displaystyle=2\cos\left(\frac{\lfloor\frac{j}{2}\rfloor}{j+1}\pi\right)-2\cos\left(\frac{\lfloor\frac{j}{2}\rfloor+1}{j+1}\pi\right)$

and
$(\zeta-1)\left(\frac{1}{\zeta}-1\right)=2-\left(\zeta+\frac{1}{\zeta}\right)=2-2\cos\left(\frac{1}{j+1}\pi\right)$,
we conclude that

$\displaystyle\Psi(W_{l,j}(x,k))$
$\displaystyle=2\sqrt{k-1}\left(\frac{2\cos\left(\frac{\lfloor\frac{j}{2}\rfloor}{j+1}\pi\right)-2\cos\left(\frac{\lfloor\frac{j}{2}\rfloor+1}{j+1}\pi\right)}{2-2\cos\left(\frac{1}{j+1}\pi\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\frac{\cos\left(\frac{\lfloor\frac{j}{2}\rfloor}{j+1}\pi\right)-\cos\left(\frac{\lfloor\frac{j}{2}\rfloor+1}{j+1}\pi\right)}{1-\cos\left(\frac{1}{j+1}\pi\right)}-1\right).$

If $j$ is odd, then $\lfloor\frac{j}{2}\rfloor=\frac{j-1}{2}$ and
$\frac{\lfloor\frac{j}{2}\rfloor+1}{j+1}\pi=\frac{\pi}{2}$, which transforms
the given expression into

$\displaystyle\Psi(W_{l,j}(x,k))$
$\displaystyle=2\sqrt{k-1}\left(\frac{\cos\left(\frac{j-1}{j+1}\cdot\frac{\pi}{2}\right)}{1-\cos\left(\frac{1}{j+1}\pi\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\frac{\sin\left(\frac{2}{j+1}\cdot\frac{\pi}{2}\right)}{2\sin^{2}\left(\frac{\pi}{2j+2}\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\frac{\sin\left(\frac{\pi}{j+1}\right)}{2\sin^{2}\left(\frac{\pi}{2j+2}\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\frac{2\sin\left(\frac{\pi}{2j+2}\right)\cos\left(\frac{\pi}{2j+2}\right)}{2\sin^{2}\left(\frac{\pi}{2j+2}\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\cot\left(\frac{\pi}{2j+2}\right)-1\right).$

If $j$ is even, then $\lfloor\frac{j}{2}\rfloor=\frac{j}{2}$ as well as
$\cos\left(\frac{\lfloor\frac{j}{2}\rfloor+1}{j+1}\pi\right)=-\cos\left(\frac{\lfloor\frac{j}{2}\rfloor}{j+1}\pi\right)$,
which gives

$\displaystyle\Psi(W_{l,j}(x,k))$
$\displaystyle=2\sqrt{k-1}\left(\frac{2\cos\left(\frac{j}{j+1}\cdot\frac{\pi}{2}\right)}{1-\cos\left(\frac{1}{j+1}\pi\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\frac{2\sin\left(\frac{1}{j+1}\cdot\frac{\pi}{2}\right)}{2\sin^{2}\left(\frac{\pi}{2j+2}\right)}-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\csc\left(\frac{\pi}{2j+2}\right)-1\right).$

∎

######  Lemma 13.

Suppose $k\geq 3$. For an even $l\geq 2$, the value $\Psi(W_{l,l+1}(x,k))$ can
be approximated via

$2\sqrt{k-1}\left(\cot\left(\frac{\pi}{2l+4}\right)-2.2\right)<\Psi(W_{l,l+1}(x,k))<2\sqrt{k-1}\
\cot\left(\frac{\pi}{2l+4}\right),$

while for an odd $l\geq 1$, the following approximation can be made:

$2\sqrt{k-1}\left(\csc\left(\frac{\pi}{2l+4}\right)-2.2\right)<\Psi(W_{l,l+1}(x,k))<2\sqrt{k-1}\
\csc\left(\frac{\pi}{2l+4}\right).$

###### Proof.

Lemma 10 directly gives us

$4\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{l+1}{2}\rfloor}\cos\left(\dfrac{h+0.5}{l+2}\pi\right)<\Psi(W_{l,l+1}(x,k))<4\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{l+1}{2}\rfloor}\cos\left(\dfrac{h-0.5}{l+2}\pi\right).$

We will denote $\zeta=e^{\frac{i\pi}{2l+4}}$. From

$\displaystyle\cos\left(\dfrac{h-0.5}{l+2}\pi\right)$
$\displaystyle=\dfrac{\zeta^{2h-1}+\zeta^{-(2h-1)}}{2}\,,$
$\displaystyle\cos\left(\dfrac{h+0.5}{l+2}\pi\right)$
$\displaystyle=\dfrac{\zeta^{2h+1}+\zeta^{-(2h+1)}}{2}\,,$

we obtain

$\displaystyle\Psi(W_{l,l+1}(x,k))$
$\displaystyle>2\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{l+1}{2}\rfloor}\left(\zeta^{2h+1}+\zeta^{-(2h+1)}\right),$
$\displaystyle\Psi(W_{l,l+1}(x,k))$
$\displaystyle<2\sqrt{k-1}\sum_{h=1}^{\lfloor\frac{l+1}{2}\rfloor}\left(\zeta^{2h-1}+\zeta^{-(2h-1)}\right).$

Since $\zeta\neq 1$, we can apply the standard formula for summing a geometric
progression in order to get

$\displaystyle\Psi(W_{l,l+1}(x,k))$ $\displaystyle<2\sqrt{k-1}\
\zeta^{1-2\lfloor\frac{l+1}{2}\rfloor}\left(1+\zeta^{2}+\zeta^{4}+\cdots+\zeta^{4\lfloor\frac{l+1}{2}\rfloor-2}\right)$
$\displaystyle=2\sqrt{k-1}\ \zeta^{1-2\lfloor\frac{l+1}{2}\rfloor}\
\dfrac{1-\zeta^{4\lfloor\frac{l+1}{2}\rfloor}}{1-\zeta^{2}}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\zeta^{1-2\lfloor\frac{l+1}{2}\rfloor}-\zeta^{2\lfloor\frac{l+1}{2}\rfloor+1}}{1-\zeta^{2}}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{(\zeta^{1-2\lfloor\frac{l+1}{2}\rfloor}-\zeta^{2\lfloor\frac{l+1}{2}\rfloor+1})(1-\zeta^{-2})}{(1-\zeta^{2})(1-\zeta^{-2})}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{(\zeta^{2\lfloor\frac{l+1}{2}\rfloor-1}+\zeta^{1-2\lfloor\frac{l+1}{2}\rfloor})-(\zeta^{2\lfloor\frac{l+1}{2}\rfloor+1}+\zeta^{-2\lfloor\frac{l+1}{2}\rfloor-1})}{2-(\zeta^{2}+\zeta^{-2})}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{2\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor-1}{2l+4}\pi\right)-2\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor+1}{2l+4}\pi\right)}{2-2\cos\left(\dfrac{1}{l+2}\pi\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor-1}{2l+4}\pi\right)-\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor+1}{2l+4}\pi\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}\,,$

as well as

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))>$
$\displaystyle>2\sqrt{k-1}\left(\zeta^{-1-2\lfloor\frac{l+1}{2}\rfloor}\left(1+\zeta^{2}+\zeta^{4}+\cdots+\zeta^{4\lfloor\frac{l+1}{2}\rfloor+2}\right)-(\zeta+\zeta^{-1})\right)$
$\displaystyle=2\sqrt{k-1}\left(\zeta^{-1-2\lfloor\frac{l+1}{2}\rfloor}\
\dfrac{1-\zeta^{4\lfloor\frac{l+1}{2}\rfloor+4}}{1-\zeta^{2}}-(\zeta+\zeta^{-1})\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\zeta^{-1-2\lfloor\frac{l+1}{2}\rfloor}-\zeta^{2\lfloor\frac{l+1}{2}\rfloor+3}}{1-\zeta^{2}}-(\zeta+\zeta^{-1})\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{(\zeta^{-1-2\lfloor\frac{l+1}{2}\rfloor}-\zeta^{2\lfloor\frac{l+1}{2}\rfloor+3})(1-\zeta^{-2})}{(1-\zeta^{2})(1-\zeta^{-2})}-(\zeta+\zeta^{-1})\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{(\zeta^{2\lfloor\frac{l+1}{2}\rfloor+1}+\zeta^{-2\lfloor\frac{l+1}{2}\rfloor-1})-(\zeta^{2\lfloor\frac{l+1}{2}\rfloor+3}+\zeta^{-2\lfloor\frac{l+1}{2}\rfloor-3})}{2-(\zeta^{2}+\zeta^{-2})}-(\zeta+\zeta^{-1})\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{2\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor+1}{2l+4}\pi\right)-2\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor+3}{2l+4}\pi\right)}{2-2\cos\left(\dfrac{1}{l+2}\pi\right)}-2\cos\left(\dfrac{\pi}{2l+4}\right)\right)$
$\displaystyle>2\sqrt{k-1}\left(\dfrac{\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor+1}{2l+4}\pi\right)-\cos\left(\dfrac{2\lfloor\frac{l+1}{2}\rfloor+3}{2l+4}\pi\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}-2\cos\left(\dfrac{\pi}{2l+4}\right)\right)\,.$

_Case_ $2\mid l$: We have $\lfloor\frac{l+1}{2}\rfloor=\frac{l}{2}$, which
allows us to conclude

$\displaystyle\Psi(W_{l,l+1}(x,k))$ $\displaystyle<2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{l-1}{2l+4}\pi\right)-\cos\left(\dfrac{l+1}{2l+4}\pi\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{l-1}{l+2}\cdot\dfrac{\pi}{2}\right)-\cos\left(\dfrac{l+1}{l+2}\cdot\dfrac{\pi}{2}\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\sin\left(\dfrac{3}{l+2}\cdot\dfrac{\pi}{2}\right)-\sin\left(\dfrac{1}{l+2}\cdot\dfrac{\pi}{2}\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{2\sin\left(\dfrac{\pi}{2l+4}\right)\cos\left(\dfrac{2\pi}{2l+4}\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{2\pi}{2l+4}\right)}{\sin\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle<2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{\pi}{2l+4}\right)}{\sin\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle=2\sqrt{k-1}\ \cot\left(\dfrac{\pi}{2l+4}\right)$

along with

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))>$
$\displaystyle>2\sqrt{k-1}\left(\dfrac{\cos\left(\dfrac{l+1}{2l+4}\pi\right)-\cos\left(\dfrac{2l+3}{2l+4}\pi\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{2\cos\left(\dfrac{l+1}{2l+4}\pi\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\sin\left(\dfrac{\pi}{2l+4}\right)}{\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle>2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2l+4}\right)-2.2\right).$

_Case_ $2\nmid l$: We have $\lfloor\frac{l+1}{2}\rfloor=\frac{l+1}{2}$, thus
obtaining

$\displaystyle\Psi(W_{l,l+1}(x,k))$ $\displaystyle<2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{l}{2l+4}\pi\right)-\cos\left(\dfrac{l+2}{2l+4}\pi\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\cos\left(\dfrac{l}{2l+4}\pi\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{\sin\left(\dfrac{2}{2l+4}\pi\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle=2\sqrt{k-1}\
\dfrac{2\sin\left(\dfrac{\pi}{2l+4}\right)\cos\left(\dfrac{\pi}{2l+4}\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}$
$\displaystyle=2\sqrt{k-1}\ \cot\left(\dfrac{\pi}{2l+4}\right)$
$\displaystyle<2\sqrt{k-1}\ \csc\left(\dfrac{\pi}{2l+4}\right)$

as well as

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))>$
$\displaystyle>2\sqrt{k-1}\left(\dfrac{\cos\left(\dfrac{l+2}{2l+4}\pi\right)-\cos\left(\dfrac{l+4}{2l+4}\pi\right)}{1-\cos\left(\dfrac{1}{l+2}\pi\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{-\cos\left(\dfrac{l+4}{2l+4}\pi\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\sin\left(\dfrac{2}{2l+4}\pi\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{2\sin\left(\dfrac{\pi}{2l+4}\right)\cos\left(\dfrac{\pi}{2l+4}\right)}{2\sin^{2}\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{\cos\left(\dfrac{\pi}{2l+4}\right)}{\sin\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\dfrac{1}{\sin\left(\dfrac{\pi}{2l+4}\right)}-\dfrac{1-\cos\left(\dfrac{\pi}{2l+4}\right)}{\sin\left(\dfrac{\pi}{2l+4}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2l+4}\right)-\dfrac{2\sin^{2}\left(\dfrac{\pi}{4l+8}\right)}{2\sin\left(\dfrac{\pi}{4l+8}\right)\cos\left(\dfrac{\pi}{4l+8}\right)}-2\cos\left(\dfrac{1}{2l+4}\pi\right)\right)$
$\displaystyle=2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2l+4}\right)-\tan\left(\dfrac{\pi}{4l+8}\right)-2\cos\left(\dfrac{\pi}{2l+4}\right)\right).$

To complete the proof, it is sufficient to show that

$\tan\left(\dfrac{\pi}{4l+8}\right)+2\cos\left(\dfrac{\pi}{2l+4}\right)\leq
2.2$

for all $l\geq 1$. For $l=1$, we have

$\displaystyle\tan\left(\dfrac{\pi}{12}\right)+2\cos\left(\dfrac{\pi}{6}\right)$
$\displaystyle=(2-\sqrt{3})+2\cdot\dfrac{\sqrt{3}}{2}=2<2.2\,.$

For $l\geq 2$, we get
$\tan\left(\dfrac{\pi}{4l+8}\right)\leq\tan\left(\dfrac{\pi}{16}\right)=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1<0.2$.
Together with $2\cos\left(\dfrac{\pi}{2l+4}\right)\leq 2$, this gives the
desired result. ∎

With the direct help of Lemma 12 and Lemma 13, we are able to formulate the
following proof.

Proof of Theorem 11. First of all, it is easy to prove Eq. (14) by taking into
consideration Theorem 9. It immediately follows that the energy of $d(l,2)$
must be equal to the sum

$E(d(l,2))=2\sum_{h=1}^{l}2\cos\left(\dfrac{h}{2l+2}\pi\right)$

Thankfully, Lemma 12 has already shown us how this exact sum can be computed
in an elegant way. In the remainder of the proof we will suppose that $k\geq
3$ and focus on proving Eq. (12) and Eq. (13).

From Eq. (4), we easily obtain

$\displaystyle E(d(l,k))=\Psi(W_{l,l+1}(x,k))$
$\displaystyle+(k-1)\Psi(W_{l,l}(x,k))$
$\displaystyle+\sum_{j=1}^{l-1}k(k-2)(k-1)^{l-1-j}\Psi(W_{l,j}(x,k))\,.$

The expression for $E(d(l,k))$ can be rewritten in the following way:

$\displaystyle E(d(l,k))=\Psi$
$\displaystyle(W_{l,l+1}(x,k))+(k-1)\Psi(W_{l,l}(x,k))$
$\displaystyle+\sum_{j=1}^{l-1}((k-1)^{2}-1)(k-1)^{l-1-j}\Psi(W_{l,j}(x,k))$
$\displaystyle=\Psi$ $\displaystyle(W_{l,l+1}(x,k))+(k-1)\Psi(W_{l,l}(x,k))$
$\displaystyle+\sum_{j=1}^{l-1}(k-1)^{l+1-j}\Psi(W_{l,j}(x,k))-\sum_{j=1}^{l-1}(k-1)^{l-1-j}\Psi(W_{l,j}(x,k))$
$\displaystyle=\Psi$ $\displaystyle(W_{l,l+1}(x,k))+(k-1)\Psi(W_{l,k}(x,k))$
$\displaystyle+\sum_{j=0}^{l-2}(k-1)^{l-j}\Psi(W_{l,j+1}(x,k))-\sum_{j=2}^{l}(k-1)^{l-j}\Psi(W_{l,j-1}(x,k))\,,$

thus giving

$\displaystyle
E(d(l,k))=\sum_{j=0}^{l}(k-1)^{l-j}\Psi(W_{l,j+1}(x,k))-\sum_{j=2}^{l}(k-1)^{l-j}\Psi(W_{l,j-1}(x,k))\,.$

We know that $\Psi(W_{l,1}(x,k))=0$ since $W_{l,1}(x,k)=x$, which means that

$\displaystyle E(d(l,k))=\sum_{j=1}^{l}(k-1)^{l-j}\Psi(W_{l,j+1}(x,k))$
$\displaystyle+(k-1)^{l}\Psi(W_{l,1}(x,k))$
$\displaystyle-\sum_{j=2}^{l}(k-1)^{l-j}\Psi(W_{l,j-1}(x,k))$
$\displaystyle=\sum_{j=1}^{l}(k-1)^{l-j}\Psi(W_{l,j+1}(x,k))$
$\displaystyle-\sum_{j=2}^{l}(k-1)^{l-j}\Psi(W_{l,j-1}(x,k))\,.$

Also, $\Psi(W_{l,0}(x,k))=0$, which implies

$\displaystyle E(d(l,k))=\sum_{j=1}^{l}(k-1)^{l-j}\Psi(W_{l,j+1}(x,k))$
$\displaystyle-\sum_{j=2}^{l}(k-1)^{l-j}\Psi(W_{l,j-1}(x,k))$
$\displaystyle-(k-1)^{l-1}\Psi(W_{l,0}(x,k))$
$\displaystyle=\sum_{j=1}^{l}(k-1)^{l-j}\Psi(W_{l,j+1}(x,k))$
$\displaystyle-\sum_{j=1}^{l}(k-1)^{l-j}\Psi(W_{l,j-1}(x,k))\,,$

from which we get

$\displaystyle E(d(l,k))$
$\displaystyle=\sum_{j=1}^{l}(k-1)^{l-j}[\Psi(W_{l,j+1}(x,k))-\Psi(W_{l,j-1}(x,k))]\,.$

The implementation of Lemma 12 gives

$\displaystyle\Psi$ $\displaystyle(W_{l,j+1}(x,k))-\Psi(W_{l,j-1}(x,k))=$
$\displaystyle=\begin{cases}2\sqrt{k-1}\left(\cot\left(\dfrac{\pi}{2j+4}\right)-1\right)-2\sqrt{k-1}\left(\cot\left(\dfrac{\pi}{2j}\right)-1\right),&2\mid
j\\\\[12.0pt]
2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2j+4}\right)-1\right)-2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2j}\right)-1\right),&2\nmid
j\end{cases}$
$\displaystyle=\begin{cases}(k-1)^{1/2}\left(2\cot\left(\dfrac{\pi}{2j+4}\right)-2\cot\left(\dfrac{\pi}{2j}\right)\right),&2\mid
j\\\\[12.0pt]
(k-1)^{1/2}\left(2\csc\left(\dfrac{\pi}{2j+4}\right)-2\csc\left(\dfrac{\pi}{2j}\right)\right),&2\nmid
j\end{cases}$ $\displaystyle=f_{j-1}(k-1)^{1/2}\,,$

for each $1\leq j\leq l-1$, which further implies

$\displaystyle E(d(l,k))$
$\displaystyle=\sum_{j=1}^{l-1}f_{j-1}(k-1)^{l+1/2-j}+[\Psi(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))]$
$\displaystyle=\sum_{j=0}^{l-2}f_{j}(k-1)^{l-1/2-j}+[\Psi(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))]\,.$

Lemma 13 helps us to approximate the final term
$\Psi(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))$. If $l$ is odd, then

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))<$
$\displaystyle<2\sqrt{k-1}\
\csc\left(\frac{\pi}{2l+4}\right)-2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2l}\right)-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\csc\left(\frac{\pi}{2l+4}\right)-\csc\left(\dfrac{\pi}{2l}\right)\right)+2\sqrt{k-1}$
$\displaystyle=f_{l-1}(k-1)^{1/2}+2(k-1)^{1/2}$

and

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))>$
$\displaystyle>2\sqrt{k-1}\left(\csc\left(\frac{\pi}{2l+4}\right)-2.2\right)-2\sqrt{k-1}\left(\csc\left(\dfrac{\pi}{2l}\right)-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\csc\left(\frac{\pi}{2l+4}\right)-\csc\left(\dfrac{\pi}{2l}\right)\right)-2.4\sqrt{k-1}$
$\displaystyle=f_{l-1}(k-1)^{1/2}-2.4(k-1)^{1/2}\,.$

If $l$ is even, then we similarly obtain

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))<$
$\displaystyle<2\sqrt{k-1}\
\cot\left(\frac{\pi}{2l+4}\right)-2\sqrt{k-1}\left(\cot\left(\dfrac{\pi}{2l}\right)-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\cot\left(\frac{\pi}{2l+4}\right)-\cot\left(\dfrac{\pi}{2l}\right)\right)+2\sqrt{k-1}$
$\displaystyle=f_{l-1}(k-1)^{1/2}+2(k-1)^{1/2}$

and

$\displaystyle\Psi$ $\displaystyle(W_{l,l+1}(x,k))-\Psi(W_{l,l-1}(x,k))>$
$\displaystyle>2\sqrt{k-1}\left(\cot\left(\frac{\pi}{2l+4}\right)-2.2\right)-2\sqrt{k-1}\left(\cot\left(\dfrac{\pi}{2l}\right)-1\right)$
$\displaystyle=2\sqrt{k-1}\left(\cot\left(\frac{\pi}{2l+4}\right)-\cot\left(\dfrac{\pi}{2l}\right)\right)-2.4\sqrt{k-1}$
$\displaystyle=f_{l-1}(k-1)^{1/2}-2.4(k-1)^{1/2}\,.$

Taking everything into consideration, we get the approximations

$\displaystyle E(d(l,k))$
$\displaystyle<\sum_{j=0}^{l-2}f_{j}(k-1)^{l-1/2-j}+f_{l-1}(k-1)^{1/2}+2(k-1)^{1/2}$
$\displaystyle=\sum_{j=0}^{l-1}f_{j}(k-1)^{l-1/2-j}+2(k-1)^{1/2}$

and

$\displaystyle E(d(l,k))$
$\displaystyle>\sum_{j=0}^{l-2}f_{j}(k-1)^{l-1/2-j}+f_{l-1}(k-1)^{1/2}-2.4(k-1)^{1/2}$
$\displaystyle=\sum_{j=0}^{l-1}f_{j}(k-1)^{l-1/2-j}-2.4(k-1)^{1/2}$

which complete the proof. ∎

Theorem 11 provides a way to approximate the energy of a dendrimer $d(l,k)$
via two expressions that resemble polynomials. To be more precise, these
expressions represent a linear combination of the power terms $(k-1)^{h-1/2},\
h\in\mathbb{N}$, where the corresponding coefficients depend solely on $l$,
not $k$. This makes it easier to analyze the asymptotic properties of
$E(d(l,k))$, leading us to the first of our two main theorems:

######  Theorem 1.

For any fixed value of $l\geq 1$, we have

$E(d(l,k))\sim 2(k-1)^{l-1/2}\qquad\mbox{as $\ k\to\infty$}\,.$ (15)

Also, for any fixed value of $k\geq 3$, we have

$E(d(l,k))\sim\mu_{k}(k-1)^{l-1/2}\qquad\mbox{as $\ l\to\infty$}\,.$ (16)

where $\mu_{k}$ is the positive real number which represents the sum of the
convergent positive series

$\sum_{j=0}^{\infty}f_{j}(k-1)^{-j}$

defined by

$f_{j}=\begin{cases}2\csc\left(\dfrac{\pi}{2j+6}\right)-2\csc\left(\dfrac{\pi}{2j+2}\right),&2\mid
j\,,\\\
2\cot\left(\dfrac{\pi}{2j+6}\right)-2\cot\left(\dfrac{\pi}{2j+2}\right),&2\nmid
j\,.\end{cases}$

In order to make the proof of Theorem 1 easier to follow, we are going to need
the following auxiliary lemma:

######  Lemma 14.

If $(a_{j})_{j\in\mathbb{N}_{0}}$ and $(b_{j})_{j\in\mathbb{N}_{0}}$ are real
sequences such that

$\displaystyle a_{j}$
$\displaystyle=2\csc\left(\dfrac{\pi}{2j+6}\right)-2\csc\left(\dfrac{\pi}{2j+2}\right),$
$\displaystyle b_{j}$
$\displaystyle=2\cot\left(\dfrac{\pi}{2j+6}\right)-2\cot\left(\dfrac{\pi}{2j+2}\right),$

for each $j\in\mathbb{N}_{0}$, then both of these sequences converge to
$\dfrac{8}{\pi}$. Also, $(a_{j})_{j\in\mathbb{N}_{0}}$ is strictly increasing,
while $(b_{j})_{j\in\mathbb{N}_{0}}$ is strictly decreasing.

###### Proof.

We have

$\displaystyle a_{j}$
$\displaystyle=2\csc\left(\frac{\pi}{2j+6}\right)-2\csc\left(\frac{\pi}{2j+2}\right)$
$\displaystyle=2\
\frac{\sin\left(\frac{\pi}{2j+2}\right)-\sin\left(\frac{\pi}{2j+6}\right)}{\sin\left(\frac{\pi}{2j+2}\right)\sin\left(\frac{\pi}{2j+6}\right)}$
$\displaystyle=2\
\frac{2\sin\left(\frac{\frac{\pi}{2j+2}-\frac{\pi}{2j+6}}{2}\right)\cos\left(\frac{\frac{\pi}{2j+2}+\frac{\pi}{2j+6}}{2}\right)}{\sin\left(\frac{\pi}{2j+2}\right)\sin\left(\frac{\pi}{2j+6}\right)}$
$\displaystyle=4\
\frac{\sin\left(\frac{\pi}{2(j+1)(j+3)}\right)\cos\left(\frac{(j+2)\pi}{2(j+1)(j+3)}\right)}{\sin\left(\frac{\pi}{2j+2}\right)\sin\left(\frac{\pi}{2j+6}\right)}$
$\displaystyle=4\
\frac{2}{\pi}\cdot\frac{\frac{\pi}{2j+2}}{\sin\left(\frac{\pi}{2j+2}\right)}\cdot\frac{\frac{\pi}{2j+6}}{\sin\left(\frac{\pi}{2j+6}\right)}\cdot\frac{\sin\left(\frac{\pi}{2(j+1)(j+3)}\right)}{\frac{\pi}{2(j+1)(j+3)}}\cdot\cos\left(\frac{(j+2)\pi}{2(j+1)(j+3)}\right)$
$\displaystyle=\frac{8}{\pi}\cdot\frac{\frac{\pi}{2j+2}}{\sin\left(\frac{\pi}{2j+2}\right)}\cdot\frac{\frac{\pi}{2j+6}}{\sin\left(\frac{\pi}{2j+6}\right)}\cdot\frac{\sin\left(\frac{\pi}{2(j+1)(j+3)}\right)}{\frac{\pi}{2(j+1)(j+3)}}\cdot\cos\left(\frac{(j+2)\pi}{2(j+1)(j+3)}\right).$

By using the well known properties of limits together with the famous formula
$\lim\limits_{x\to 0}\dfrac{\sin x}{x}=1$, it becomes obvious that
$\lim\limits_{j\to\infty}a_{j}=\dfrac{8}{\pi}$. Similarly, we obtain

$\displaystyle b_{j}$
$\displaystyle=2\cot\left(\frac{\pi}{2j+6}\right)-2\cot\left(\frac{\pi}{2j+2}\right)$
$\displaystyle=2\
\frac{\sin\left(\frac{\pi}{2j+2}\right)\cos\left(\frac{\pi}{2j+6}\right)-\sin\left(\frac{\pi}{2j+6}\right)\cos\left(\frac{\pi}{2j+2}\right)}{\sin\left(\frac{\pi}{2j+2}\right)\sin\left(\frac{\pi}{2j+6}\right)}$
$\displaystyle=2\
\frac{\sin\left(\frac{\pi}{2j+2}-\frac{\pi}{2j+6}\right)}{\sin\left(\frac{\pi}{2j+2}\right)\sin\left(\frac{\pi}{2j+6}\right)}$
$\displaystyle=2\
\frac{\sin\left(\frac{\pi}{(j+1)(j+3)}\right)}{\sin\left(\frac{\pi}{2j+2}\right)\sin\left(\frac{\pi}{2j+6}\right)}$
$\displaystyle=2\
\frac{4}{\pi}\cdot\frac{\frac{\pi}{2j+2}}{\sin\left(\frac{\pi}{2j+2}\right)}\cdot\frac{\frac{\pi}{2j+6}}{\sin\left(\frac{\pi}{2j+6}\right)}\cdot\frac{\sin\left(\frac{\pi}{(j+1)(j+3)}\right)}{\frac{\pi}{(j+1)(j+3)}}$
$\displaystyle=\frac{8}{\pi}\cdot\frac{\frac{\pi}{2j+2}}{\sin\left(\frac{\pi}{2j+2}\right)}\cdot\frac{\frac{\pi}{2j+6}}{\sin\left(\frac{\pi}{2j+6}\right)}\cdot\frac{\sin\left(\frac{\pi}{(j+1)(j+3)}\right)}{\frac{\pi}{(j+1)(j+3)}}\,,$

which implies $\lim\limits_{j\to\infty}b_{j}=\dfrac{8}{\pi}$.

Let us define the function $y_{1}(x)=\csc\left(\dfrac{\pi}{x}\right)$ on
$[2,+\infty)$. This function is obviously twice differentiable everywhere. We
know that

$\displaystyle a_{j}-a_{j-1}$
$\displaystyle=2\csc\left(\dfrac{\pi}{2j+6}\right)-2\csc\left(\dfrac{\pi}{2j+2}\right)-2\csc\left(\dfrac{\pi}{2j+4}\right)+2\csc\left(\dfrac{\pi}{2j}\right)$
$\displaystyle=2(y_{1}(2j+6)-y_{1}(2j+4))-2(y_{1}(2j+2)-y_{1}(2j))$

for each $j\geq 1$. Since the function $y_{1}$ is differentiable everywhere,
we can use the mean value theorem to get

$\displaystyle a_{j}-a_{j-1}$
$\displaystyle=2((2j+6)-(2j+4))y_{1}^{\prime}(\xi_{1})-2((2j+2)-2j)y_{1}^{\prime}(\xi_{2})$
$\displaystyle=4y_{1}^{\prime}(\xi_{1})-4y_{1}^{\prime}(\xi_{2})$

for some $\xi_{1}\in(2j+4,2j+6),\ \xi_{2}\in(2j,2j+2)$. Since
$\xi_{1}>\xi_{2}>2$, in order to prove that $(a_{j})_{j\in\mathbb{N}_{0}}$ is
strictly increasing, it is sufficient to show that $y_{1}^{\prime\prime}(x)$
is positive on $(2,+\infty)$. We compute

$\displaystyle y_{1}^{\prime}(x)$
$\displaystyle=-\dfrac{\cos\left(\dfrac{\pi}{x}\right)}{\sin^{2}\left(\dfrac{\pi}{x}\right)}\cdot\left(-\dfrac{\pi}{x^{2}}\right)=\dfrac{\cos\left(\dfrac{\pi}{x}\right)}{\sin^{2}\left(\dfrac{\pi}{x}\right)}\cdot\dfrac{\pi}{x^{2}}=\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right),$

which further gives

$\displaystyle y_{1}^{\prime\prime}(x)$
$\displaystyle=\left(\dfrac{\pi}{x^{2}}\right)^{\prime}\csc\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right)+\dfrac{\pi}{x^{2}}\csc^{\prime}\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right)+\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\cot^{\prime}\left(\dfrac{\pi}{x}\right)$
$\displaystyle=-\dfrac{2\pi}{x^{3}}\csc\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right)+\dfrac{\pi}{x^{2}}\left(\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right)\right)\cot\left(\dfrac{\pi}{x}\right)$
$\displaystyle\qquad\qquad+\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\left(-\csc^{2}\left(\dfrac{\pi}{x}\right)\right)\cdot\left(-\dfrac{\pi}{x^{2}}\right)$
$\displaystyle=-\dfrac{2\pi}{x^{3}}\csc\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right)+\dfrac{\pi^{2}}{x^{4}}\left(\csc\left(\dfrac{\pi}{x}\right)\cot^{2}\left(\dfrac{\pi}{x}\right)+\csc^{3}\left(\dfrac{\pi}{x}\right)\right)$
$\displaystyle=\dfrac{\pi}{x^{4}}\csc\left(\dfrac{\pi}{x}\right)\left(-2x\cot\left(\dfrac{\pi}{x}\right)+\pi\cot^{2}\left(\dfrac{\pi}{x}\right)+\pi\csc^{2}\left(\dfrac{\pi}{x}\right)\right).$

Here, it is possible to conclude that for $x\in(2,+\infty)$

$\displaystyle y_{1}^{\prime\prime}(x)>0$
$\displaystyle\iff-2x\cot\left(\dfrac{\pi}{x}\right)+\pi\cot^{2}\left(\dfrac{\pi}{x}\right)+\pi\csc^{2}\left(\dfrac{\pi}{x}\right)>0$
$\displaystyle\iff\cot^{2}\left(\dfrac{\pi}{x}\right)+\csc^{2}\left(\dfrac{\pi}{x}\right)>\dfrac{2x}{\pi}\cot\left(\dfrac{\pi}{x}\right)$
$\displaystyle\iff\cos^{2}\left(\dfrac{\pi}{x}\right)+1>\dfrac{2x}{\pi}\sin\left(\dfrac{\pi}{x}\right)\cos\left(\dfrac{\pi}{x}\right)$
$\displaystyle\iff\cos^{2}\left(\dfrac{\pi}{x}\right)+1>\dfrac{2\sin\left(\dfrac{\pi}{x}\right)\cos\left(\dfrac{\pi}{x}\right)}{\dfrac{\pi}{x}}.$

Hence, in order to prove that $y_{1}^{\prime\prime}(x)>0$ for all
$x\in(2,+\infty)$, it is sufficient to show that
$\cos^{2}(\theta)+1>\dfrac{2\sin\theta\cos\theta}{\theta}$ for each
$\theta\in(0,\pi/2)$. However, we know that

$\displaystyle\cos^{2}\theta+1$
$\displaystyle>\dfrac{2\sin\theta\cos\theta}{\theta}$ $\displaystyle\iff$
$\displaystyle\cos^{2}\theta-2\cos\theta+1$
$\displaystyle>\dfrac{2\sin\theta\cos\theta}{\theta}-2\cos\theta$
$\displaystyle\iff$ $\displaystyle(\cos\theta-1)^{2}$
$\displaystyle>2\cos\theta\cdot\left(\dfrac{\sin\theta}{\theta}-1\right),$

which obviously must hold for all $\theta\in(0,\pi/2)$, since
$(\cos\theta-1)^{2}>0$, while $\dfrac{\sin\theta}{\theta}-1<0$ due to
$0<\sin\theta<\theta$, which is known to hold on $(0,\pi/2)$.

Now we define the function $y_{2}(x)=\cot\left(\dfrac{\pi}{x}\right)$ on
$[2,+\infty)$. This function is also twice differentiable everywhere. We have

$\displaystyle b_{j}-b_{j-1}$
$\displaystyle=2\cot\left(\dfrac{\pi}{2j+6}\right)-2\cot\left(\dfrac{\pi}{2j+2}\right)-2\cot\left(\dfrac{\pi}{2j+4}\right)+2\cot\left(\dfrac{\pi}{2j}\right)$
$\displaystyle=2(y_{2}(2j+6)-y_{2}(2j+4))-2(y_{2}(2j+2)-y_{2}(2j))$

for each $j\geq 1$. Due to the differentiability of $y_{2}$ on its entire
domain, we can implement the mean value theorem to obtain

$\displaystyle b_{j}-b_{j-1}$
$\displaystyle=2((2j+6)-(2j+4))y_{2}^{\prime}(\eta_{1})-2((2j+2)-2j)y_{2}^{\prime}(\eta_{2})$
$\displaystyle=4y_{2}^{\prime}(\eta_{1})-4y_{2}^{\prime}(\eta_{2})$

for some $\eta_{1}\in(2j+4,2j+6),\ \eta_{2}\in(2j,2j+2)$. Because of
$\eta_{1}>\eta_{2}>2$, in order to prove that $(b_{j})_{j\in\mathbb{N}_{0}}$
is strictly decreasing, it is sufficient to show that
$y_{2}^{\prime\prime}(x)$ is negative on $(2,+\infty)$. We compute

$\displaystyle y_{2}^{\prime}(x)$
$\displaystyle=-\csc^{2}\left(\dfrac{\pi}{x}\right)\cdot\left(-\dfrac{\pi}{x^{2}}\right)=\dfrac{\pi}{x^{2}}\csc^{2}\left(\dfrac{\pi}{x}\right)$

along with

$\displaystyle y_{2}^{\prime\prime}(x)$
$\displaystyle=\left(\dfrac{\pi}{x^{2}}\right)^{\prime}\csc^{2}\left(\dfrac{\pi}{x}\right)+2\
\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\csc^{\prime}\left(\dfrac{\pi}{x}\right)$
$\displaystyle=-\dfrac{2\pi}{x^{3}}\csc^{2}\left(\dfrac{\pi}{x}\right)+2\
\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\cdot\left(\dfrac{\pi}{x^{2}}\csc\left(\dfrac{\pi}{x}\right)\cot\left(\dfrac{\pi}{x}\right)\right)$
$\displaystyle=\dfrac{\pi}{x^{4}}\csc^{2}\left(\dfrac{\pi}{x}\right)\left(-2x+2\pi\cot\left(\dfrac{\pi}{x}\right)\right).$

Thus, for $x\in(2,+\infty)$ we have

$\displaystyle y_{2}^{\prime\prime}(x)<0$
$\displaystyle\iff-2x+2\pi\cot\left(\dfrac{\pi}{x}\right)<0$
$\displaystyle\iff\dfrac{\pi}{x}\cot\left(\dfrac{\pi}{x}\right)<1$
$\displaystyle\iff\dfrac{\pi}{x}<\tan\left(\dfrac{\pi}{x}\right)$

However, it is known that the inequality $\tan\theta>\theta$ holds on
$(0,\pi/2)$, which proves that the sequence $(b_{j})_{j\in\mathbb{N}_{0}}$
must be strictly decreasing. ∎

We are now in the position to prove Theorem 1 by extensively relying on the
sequences $(a_{j})_{j\in\mathbb{N}_{0}}$ and $(b_{j})_{j\in\mathbb{N}_{0}}$
defined in Lemma 14, as well their properties which we have shown.

Proof of Theorem 1. First of all, we are going to prove Eq. (15). Directly
from Theorem 11, we obtain:

$\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}<\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}+2(k-1)^{-(l-1)}\,,$
(17)

as well as

$\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}>\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}-2.4(k-1)^{-(l-1)}\,.$
(18)

Provided $l\geq 2$, it is clear that when the $l$ variable is fixed and
$k\to\infty$, then both the right-hand side in Eq. (17) and the right-hand
side in Eq. (18) tend to $f_{0}$. By the squeeze theorem, we get that

$\displaystyle\lim_{k\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}=f_{0}\,.$

Taking into consideration that $f_{0}=a_{0}=2$, Eq. (15) is proven for $l\geq
2$. For $l=1$, the formula needs to be proven directly. From Eq. (5) we see
that $E(d(1,k))=2\sqrt{k}$. Having this in mind, it is obvious that

$\displaystyle\lim_{k\to\infty}\dfrac{E(d(1,k))}{(k-1)^{1/2}}=2\,,$

which completes the proof of Eq. (15).

The next step is to prove that the positive series

$\sum_{j=0}^{\infty}f_{j}(k-1)^{-j}$

is convergent. From Lemma 14, it is clear that all the $a_{j}$ must be smaller
than $\dfrac{8}{\pi}$, while all the $b_{j}$ must be greater than
$\dfrac{8}{\pi}$. Given the fact that $f_{j}=a_{j}$ when $j$ is even and
$f_{j}=b_{j}$ when $j$ is odd, it is easy to establish that

$\sup_{j\in\mathbb{N}_{0}}f_{j}=f_{1}$

In other words, if we denote $F_{j}=\sum\limits_{h=0}^{j}f_{h}(k-1)^{-h}$,
then

$F_{j}\leq\sum_{h=0}^{j}f_{1}(k-1)^{-h}$

for each $j\in\mathbb{N}_{0}$. Subsequently,

$\displaystyle F_{j}$ $\displaystyle\leq f_{1}\sum_{h=0}^{j}(k-1)^{-h}$
$\displaystyle=f_{1}\ \dfrac{1-(k-1)^{-j-1}}{1-(k-1)^{-1}}$
$\displaystyle<\dfrac{f_{1}}{1-(k-1)^{-1}}$

which means that the positive series
$\sum\limits_{j=0}^{\infty}f_{j}(k-1)^{-j}$ must have a sequence of partial
sums which is bounded. This implies that the series is convergent.

Finally, we are going to prove Eq. (16). Suppose that the variable $k\geq 3$
is fixed. From Eq. (17) we get

$\displaystyle\limsup_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}\leq\limsup_{l\to\infty}\left(\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}+2(k-1)^{-(l-1)}\right).$

Having proven the convergence of $\sum\limits_{j=0}^{\infty}f_{j}(k-1)^{-j}$,
we know that its sum is a positive real number $\mu_{k}$. From

$\displaystyle\lim\limits_{l\to\infty}\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}$
$\displaystyle=\mu_{k}\,,$
$\displaystyle\lim\limits_{l\to\infty}2(k-1)^{-(l-1)}$ $\displaystyle=0\,,$

we obtain

$\limsup_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}\leq\mu_{k}\,.$

Similarly, from Eq. (18) we have

$\displaystyle\liminf_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}\geq\liminf_{l\to\infty}\left(\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}-2.4(k-1)^{-(l-1)}\right),$

which ultimately gives

$\liminf_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}\geq\mu_{k}\,.$

From

$\mu_{k}\leq\liminf_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}\leq\limsup_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}\leq\mu_{k}$

we get $\lim\limits_{l\to\infty}\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}=\mu_{k}$,
which proves Eq. (16), as desired. ∎

We will end this paper by proving our second main result, along with one of
its direct corollaries.

######  Theorem 2.

For a given dendrimer $d(l,k)$, where $k\geq 3$ and $l\geq 2$, we have

$\displaystyle E(d(l,k))$
$\displaystyle<(k-1)^{l-1/2}\left(2+\dfrac{0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{k-1}\right),$
(19) $\displaystyle E(d(l,k))$
$\displaystyle>(k-1)^{l-1/2}\left(2+\dfrac{2\sqrt{2}}{k-1}\right).$ (20)

###### Proof.

For $l=2$, we are going to prove that both Eq. (19) and Eq. (20) hold
directly. From Eq. (6), we conclude that
$E(d(2,k))=2(k-1)^{3/2}+2\sqrt{2k-1}$, which gives us

$\displaystyle\dfrac{E(d(2,k))}{(k-1)^{3/2}}$
$\displaystyle=2+2(k-1)^{-3/2}\sqrt{2k-1}$
$\displaystyle=2+\dfrac{2}{k-1}\sqrt{\dfrac{2k-1}{k-1}}$
$\displaystyle=2+\dfrac{2}{k-1}\sqrt{2+\dfrac{1}{k-1}}$

We obviously have $\sqrt{2+\dfrac{1}{k-1}}>\sqrt{2}$, which proves Eq. (20).
However, due to $k\geq 3$, we also have
$\sqrt{2+\dfrac{1}{k-1}}\leq\sqrt{\dfrac{5}{2}}<\sqrt{5}<0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}$,
thereby proving Eq. (19) as well.

For $l=3$, the expressions will again be proven directly. From Theorem 11, we
get

$\displaystyle E(d(3,k))$
$\displaystyle<\sum_{j=0}^{2}f_{j}(k-1)^{5/2-j}+2(k-1)^{1/2}\,,$
$\displaystyle E(d(3,k))$
$\displaystyle>\sum_{j=0}^{2}f_{j}(k-1)^{5/2-j}-2.4(k-1)^{1/2}\,,$

where

$\displaystyle f_{0}$
$\displaystyle=2\csc\left(\dfrac{\pi}{6}\right)-2\csc\left(\dfrac{\pi}{2}\right)=2\,,$
$\displaystyle f_{1}$
$\displaystyle=2\cot\left(\dfrac{\pi}{8}\right)-2\cot\left(\dfrac{\pi}{4}\right)=2\sqrt{2}\,,$
$\displaystyle f_{2}$
$\displaystyle=2\csc\left(\dfrac{\pi}{10}\right)-2\csc\left(\dfrac{\pi}{6}\right)=2\sqrt{5}-2\,.$

Hence, we obtain

$\displaystyle\dfrac{E(d(3,k))}{(k-1)^{5/2}}$
$\displaystyle<2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2+2}{(k-1)^{2}}\,,$
$\displaystyle\dfrac{E(d(3,k))}{(k-1)^{5/2}}$
$\displaystyle>2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2-2.4}{(k-1)^{2}}\,.$

We know that $2\sqrt{5}>4.4=2+2.4$, which means that Eq. (20) must hold for
$l=3$. On the other hand, we get

$\displaystyle\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2+2}{(k-1)^{2}}$
$\displaystyle=\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}}{(k-1)^{2}}$
$\displaystyle=\dfrac{2\sqrt{2}}{k-1}+\dfrac{1}{k-1}\cdot\dfrac{2\sqrt{5}}{k-1}$
$\displaystyle\leq\dfrac{2\sqrt{2}}{k-1}+\dfrac{1}{2}\cdot\dfrac{2\sqrt{5}}{k-1}$
$\displaystyle=\dfrac{2\sqrt{2}+\sqrt{5}}{k-1}\,,$

which proves Eq. (19), given the fact that
$2\sqrt{2}+\sqrt{5}<0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}$.

Now, suppose that $l\geq 4$. By using Theorem 11 once again, we directly
conclude that

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle<\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}+2(k-1)^{-(l-1)}\,,$ (21)
$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle>\sum_{j=0}^{l-1}f_{j}(k-1)^{-j}-2.4(k-1)^{-(l-1)}\,.$ (22)

From Eq. (22), we further obtain

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle>\sum_{j=0}^{2}f_{j}(k-1)^{-j}-2.4(k-1)^{-(l-1)}$
$\displaystyle=2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}-2.4(k-1)^{-(l-1)}$
$\displaystyle>2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}-2.4(k-1)^{-2}$
$\displaystyle=2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-4.4}{(k-1)^{2}}\,,$

which completes our entire proof of Eq. (20), given the fact that
$2\sqrt{5}>4.4$.

From Eq. (21), we have

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle<\sum_{j=0}^{+\infty}f_{j}(k-1)^{-j}+2(k-1)^{-(l-1)}$
$\displaystyle\leq\sum_{j=0}^{+\infty}f_{j}(k-1)^{-j}+2(k-1)^{-3}$
$\displaystyle<2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}+\dfrac{2}{(k-1)^{3}}+\sum_{j=3}^{\infty}f_{j}(k-1)^{-j}\,.$

Here, it is important to notice that $\sup\limits_{j\geq 3}f_{j}=f_{3}$ due to
the results obtained in Lemma 14. This implies

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle<2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}+\dfrac{2}{(k-1)^{3}}+\sum_{j=3}^{\infty}f_{3}(k-1)^{-j}$
$\displaystyle=2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}+\dfrac{2}{(k-1)^{3}}+\dfrac{f_{3}}{(k-1)^{3}}\sum_{j=0}^{\infty}(k-1)^{-j}$
$\displaystyle=2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}+\dfrac{2}{(k-1)^{3}}+\dfrac{f_{3}}{(k-1)^{3}}\cdot\dfrac{1}{1-\frac{1}{k-1}}$
$\displaystyle=2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{2\sqrt{5}-2}{(k-1)^{2}}+\dfrac{2}{(k-1)^{3}}+\dfrac{f_{3}}{(k-2)(k-1)^{2}}\,.$

It is clear that

$\displaystyle\dfrac{1}{(k-1)^{2}}$
$\displaystyle\leq\dfrac{1}{2}\cdot\dfrac{1}{k-1}\,,$
$\displaystyle\dfrac{1}{(k-1)^{3}}$
$\displaystyle\leq\dfrac{1}{4}\cdot\dfrac{1}{k-1}\,,$
$\displaystyle\dfrac{1}{(k-2)(k-1)^{2}}$
$\displaystyle\leq\dfrac{1}{2}\cdot\dfrac{1}{k-1}\,,$

as well was

$\displaystyle f_{3}$
$\displaystyle=2\cot\left(\dfrac{\pi}{12}\right)-2\cot\left(\dfrac{\pi}{8}\right)$
$\displaystyle=2(2+\sqrt{3})-2(\sqrt{2}+1)$
$\displaystyle=2-2\sqrt{2}+2\sqrt{3}\,.$

We then obtain

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle<2+\dfrac{2\sqrt{2}}{k-1}+\dfrac{\sqrt{5}-1}{k-1}+\dfrac{0.5}{k-1}+\dfrac{1-\sqrt{2}+\sqrt{3}}{k-1}$
$\displaystyle=2+\dfrac{2\sqrt{2}+\sqrt{5}-1+0.5+1-\sqrt{2}+\sqrt{3}}{k-1}$
$\displaystyle=2+\dfrac{0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{k-1}\,,$

thereby proving Eq. (19) for $l\geq 4$. ∎

######  Corollary 15.

We have

$\displaystyle E(d(l,k))=\Theta((k-1)^{l-1/2})\qquad\mbox{as $(l,k)\to\infty$
over $\mathbb{N}\times(\mathbb{N}\setminus\\{1,2\\})$}\,.$

###### Proof.

For $l=1$ and $k\geq 3$, we have $E(d(2,k))=2\sqrt{k}$, making it
straightforward to conclude that

$\displaystyle\dfrac{E(d(2,k))}{(k-1)^{1/2}}$ $\displaystyle\leq
2\sqrt{\dfrac{3}{2}}\,,$ $\displaystyle\dfrac{E(d(2,k))}{(k-1)^{1/2}}$
$\displaystyle>2\,,$

i.e. $\dfrac{E(d(2,k))}{(k-1)^{1/2}}\in\left(2,\sqrt{6}\right]$. For $l\geq 2$
and $k\geq 3$ we can use Theorem 2 to obtain

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle<2+\dfrac{0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{k-1}\,,$
$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle>2+\dfrac{2\sqrt{2}}{k-1}\,,$

which further implies

$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$
$\displaystyle<2+\dfrac{0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}\,,$
$\displaystyle\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}$ $\displaystyle>2\,.$

Since $2+\dfrac{0.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}>\sqrt{6}$, we conclude that
for all $l\geq 1$ and $k\geq 3$, the given inequality must hold:

$2<\dfrac{E(d(l,k))}{(k-1)^{l-1/2}}<\dfrac{4.5+\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}\,.$

The theorem statement follows directly. ∎

## Conflict of interest

The authors declare that they have no conflict of interest.

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