superstrings, SO(32), E${}_{8}\times$E8 heterotic superstrings, M-theory,
F-theory. In F-theory the 5-brane represents the SL(5) U-duality symmetry
[46]. Recently it was shown that current algebras of 5-branes are preserved
under the S and T-duality transformations with renaming the spacetime
coordinates, where 5-branes include the NS5-brane, the D5-brane, KK5-branes
and exotic 5-branes in 32-supersymmetric string theories [54]. 5-brane may
give a clue of duality web including 16 supersymmetric string theories.

Many interesting topics are unsolved such as supersymmetric actions of
F-theory and M-theory, first quantization of branes and spectrum, amplitudes,
and duality web including 16-supersymmetric theories.

## Acknowledgements

We would like to thank Di Wang for useful discussions. M.H. would like to
thank the Simons Center for Geometry and Physics for hospitality during “the
2019 Summer Simons workshop in Mathematics and Physics” where this work has
been developed. W.S. is supported by NSF grant PHY-1915093.

##  Appendix A Indices

Indices are summarized as follows.

| number of d.o.f. |  curved |  flat
---|---|---|---
F-theory |  |  |
spacetime |  |  |
SO(5,5) spinor | 16 | $\mu,\nu,\cdots$ | $\alpha,\beta,\cdots$
SO(6,6) spinor | 32 | $\underline{\mu},\underline{\nu},\cdots$ | $\underline{\alpha},\underline{\beta},\cdots$
GL(5) tensor | 16 | $M,N,\cdots$ | $A,B,\cdots$
GL(6) tensor | 32 | $\underline{M},\underline{N},\cdots$ | $\underline{A},\underline{B},\cdots$
worldvolume |  |  |
SO(5,5) vector | 10 | $\underline{m},\underline{n},\cdots$ | $\underline{a},\underline{b},\cdots$
SO(6,6) vector | 12 | $\underline{\hat{m}},\underline{\hat{n}},\cdots$ | $\underline{\hat{a}},\underline{\hat{b}},\cdots$
GL(5) vector | 5 | $m,n,\cdots$ | $a,b,\cdots$
GL(6) vector | 6 | $\hat{m},\hat{n},\cdots$ | $\hat{a},\hat{b},\cdots$
T-theory |  |  |
(only $\S$4-1) |  |  |
spacetime |  |  |
O($D,D$) vector | $2D$ | $M,N,\cdots$ | $A,B,\cdots$
left-handed | $D$ | $\overline{M},\overline{N},\cdots$ | $\overline{A},\overline{B},\cdots$
right-handed | $D$ | $\underline{M},\underline{N},\cdots$ | $\underline{A},\underline{B},\cdots$
worldvolume |  |  |
SO(1,1) vector | 2 | ${m},{n},\cdots$ | ${a},{b},\cdots$

##  Appendix B Brackets

In the F-theory spacetime the Lie derivative is modified in such a way that it
is the SO(5,5) U-duality symmetry covariant. We compute commutators in the
SO(5,5) spinor representation which is easier than the GL(5) representation.
For vector functions $V_{i}{}^{\mu}(X)$ with $i=1,2$ in the 16-dimensional
spacetime a commutator brackets of these vectors is given by

$\displaystyle\left[V_{1}^{\mu}\hbox{\large$\vartriangleright$}_{\mu}(\sigma),V_{2}^{\nu}\hbox{\large$\vartriangleright$}_{\nu}(\sigma^{\prime})\right]$
$\displaystyle=$ $\displaystyle
2i\left(\frac{1-K}{2}\Phi_{(12)}^{\underline{m}}(\sigma)+\frac{1+K}{2}\Phi_{(12)}^{\underline{m}}(\sigma^{\prime})\right)\partial_{\underline{m}}\delta(\sigma-\sigma^{\prime})$
$\displaystyle-i\left(V_{1}^{\nu}\partial_{\nu}V_{2}^{\mu}-V_{2}^{\nu}\partial_{\nu}V_{1}^{\mu}\right)\hbox{\large$\vartriangleright$}_{\mu}\delta(\sigma-\sigma^{\prime})$
$\displaystyle+i\left(\frac{1-K}{2}V_{1}\gamma^{\underline{m}}\partial_{\mu}V_{2}-\frac{1+K}{2}\partial_{\mu}V_{1}\gamma^{\underline{m}}V_{2}\right)(\gamma_{\underline{m}}\hbox{\large$\vartriangleright$})^{\mu}\delta(\sigma-\sigma^{\prime})~{}~{}$
$\displaystyle\Phi_{(12)}^{\underline{m}}$ $\displaystyle=$ $\displaystyle
V_{1}{}^{\mu}\gamma^{\underline{m}}{}_{\mu\nu}V_{2}{}^{\nu}~{}~{}~{}.$

The exceptional Courant bracket is given by $K=0$ as

$\displaystyle\left[V_{1}^{\mu}\hbox{\large$\vartriangleright$}_{\mu}(\sigma),V_{2}^{\nu}\hbox{\large$\vartriangleright$}_{\nu}(\sigma^{\prime})\right]$
$\displaystyle=$
$\displaystyle{i}\left(\Phi_{(12)}^{\underline{m}}(\sigma)+\Phi_{(12)}^{\underline{m}}(\sigma^{\prime})\right)\partial_{\underline{m}}\delta(\sigma-\sigma^{\prime})-iV_{[12]}{}^{\mu}\hbox{\large$\vartriangleright$}_{\mu}\delta(\sigma-\sigma^{\prime})$
$\displaystyle V_{[12]}{}^{\mu}$ $\displaystyle=$ $\displaystyle
V_{[1|}^{\nu}\partial_{\nu}V_{|2]}^{\mu}-\frac{1}{2}(V_{[1|}{}^{\rho}\gamma^{\underline{m}}{}_{\rho\lambda}\partial_{\nu}V_{|2]}{}^{\lambda})\gamma_{\underline{m}}{}^{\nu\mu}~{}~{},$
(B.2)

while the exceptional Dorfman bracket is given by $K=1$ as

$\displaystyle\left[V_{1}^{\mu}\hbox{\large$\vartriangleright$}_{\mu}(\sigma),V_{2}^{\nu}\hbox{\large$\vartriangleright$}_{\nu}(\sigma^{\prime})\right]$
$\displaystyle=$ $\displaystyle
2i\Phi_{(12)}^{\underline{m}}(\sigma^{\prime})\partial_{\underline{m}}\delta(\sigma-\sigma^{\prime})$
$\displaystyle-i\left(V_{[1|}^{\nu}\partial_{\nu}V_{|2]}^{\mu}+(\partial_{\nu}V_{1}\gamma^{\underline{m}}V_{2})\gamma_{\underline{m}}{}^{\nu\mu}\right)\hbox{\large$\vartriangleright$}_{\mu}\delta(\sigma-\sigma^{\prime})~{}~{}.$

##  Appendix C 11-dimensional tensor representation

The 16 component SO(5,5) spinor current is decomposed under the GL(5) as
$16\to 5\oplus 10\oplus 1$. We present the current algebras preserving the
full tensor indices such as $\hbox{\large$\vartriangleright$}^{m_{1}\cdots
m_{5}}$ with $m=1,\cdots,5$ in order to manifest the 11-dimensional
supergravity background. The SO(5,5) current for the M5-brane is obtained as
$\hbox{\large$\vartriangleright$}_{M}=(\hbox{\large$\vartriangleright$}_{m},\hbox{\large$\vartriangleright$}^{m_{1}m_{2}},\hbox{\large$\vartriangleright$}^{m_{1}\cdots
m_{5}})$ . The F-theory SO(5,5) current algebras give in (3.13) are rewritten
in terms of the GL(5) tensors as

$\displaystyle\left\\{\begin{array}[]{rcl}\left[\hbox{\large$\vartriangleright$}_{m}(\sigma),\hbox{\large$\vartriangleright$}^{n_{1}n_{2}}(\sigma^{\prime})\right]&=&2i\delta_{m}^{[n_{1}}\partial^{n_{2}]}\delta(\sigma-\sigma^{\prime})\\\
\left[\hbox{\large$\vartriangleright$}_{m}(\sigma),\hbox{\large$\vartriangleright$}^{n_{1}\cdots
n_{5}}(\sigma^{\prime})\right]&=&\frac{2i}{4!}\delta_{m}^{[n_{1}}\partial^{n_{2}\cdots
n_{5}]}\delta(\sigma-\sigma^{\prime})\\\
\left[\hbox{\large$\vartriangleright$}^{m_{1}m_{2}}(\sigma),\hbox{\large$\vartriangleright$}^{n_{3}n_{4}}(\sigma^{\prime})\right]&=&2i\partial^{m_{1}m_{2}n_{3}n_{4}}\delta(\sigma-\sigma^{\prime})\end{array}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left[{\hbox{\large$\vartriangleright$}}_{M}(\sigma),\tilde{\hbox{\large$\vartriangleright$}}_{N}(\sigma^{\prime})\right]~{}=~{}0$
$\displaystyle\left\\{\begin{array}[]{rcl}\left[\tilde{\hbox{\large$\vartriangleright$}}_{m}(\sigma),\tilde{\hbox{\large$\vartriangleright$}}^{n_{1}n_{2}}(\sigma^{\prime})\right]&=&-2i\delta_{m}^{[n_{1}}\partial^{n_{2}]}\delta(\sigma-\sigma^{\prime})\\\
\left[\tilde{\hbox{\large$\vartriangleright$}}_{m}(\sigma),\tilde{\hbox{\large$\vartriangleright$}}^{n_{1}\cdots
n_{5}}(\sigma^{\prime})\right]&=&-\frac{2i}{4!}\delta_{m}^{[n_{1}}\partial^{n_{2}\cdots
n_{5}]}\delta(\sigma-\sigma^{\prime})\\\
\left[\tilde{\hbox{\large$\vartriangleright$}}^{m_{1}m_{2}}(\sigma),\tilde{\hbox{\large$\vartriangleright$}}^{n_{3}n_{4}}(\sigma^{\prime})\right]&=&-2i\partial^{m_{1}m_{2}n_{3}n_{4}}\delta(\sigma-\sigma^{\prime})\end{array}\right.$

where the $\delta(\sigma)$ stands for the 10-dimensional worldvolume function
$\delta^{(10)}(\sigma-\sigma^{\prime})$. The bosonic coordinates,
$X^{M}=(X^{m},X_{m_{1}m_{2}},X_{m_{1}\cdots m_{5}})$ and
$P_{M}=(P_{m},P^{m_{1}m_{2}},P^{m_{1}\cdots m_{5}})$, satisfy the following
canonical commutators

$\displaystyle\left\\{\begin{array}[]{rcl}\left[P_{m}(\sigma),X^{n}(\sigma)\right]&=&\frac{1}{i}\delta_{m}^{n}\delta(\sigma-\sigma^{\prime})\\\
\left[P^{m_{1}m_{2}}(\sigma),X_{n_{1}n_{2}}(\sigma)\right]&=&\frac{1}{i}\delta_{m_{1}}^{[n_{1}}\delta_{m_{2}}^{n_{2}]}\delta(\sigma-\sigma^{\prime})\\\
\left[P^{m_{1}\cdots m_{5}}(\sigma),X_{n_{1}\cdots
n_{5}}(\sigma)\right]&=&\frac{1}{i}\delta_{m_{1}}^{[n_{1}}\cdots\delta_{m_{5}}^{n_{5}]}\delta(\sigma-\sigma^{\prime})\end{array}\right.~{}~{}~{}.$
(C.8)

The covariant derivatives, which are selfdual currents, are given as

$\displaystyle\left\\{\begin{array}[]{rcl}\hbox{\large$\vartriangleright$}_{m}&=&P_{m}+\partial^{n}X_{mn}+\frac{1}{4!}\partial^{m_{1}\cdots
m_{4}}X_{mm_{1}\cdots m_{4}}\\\
\hbox{\large$\vartriangleright$}^{m_{1}m_{2}}&=&P^{m_{1}m_{2}}-\partial^{[m_{1}}X^{m_{2}]}+\frac{1}{2}\partial^{m_{1}\cdots
m_{4}}X_{m_{3}m_{4}}\\\ \hbox{\large$\vartriangleright$}^{m_{1}\cdots
m_{5}}&=&P^{m_{1}\cdots m_{5}}+\frac{1}{4!}\partial^{[m_{1}\cdots
m_{4}}X^{m_{5}]}\end{array}\right.~{}~{}~{},$ (C.12)

and the symmetry generators, which are anti-selfdual currents, are given as

$\displaystyle\left\\{\begin{array}[]{rcl}\tilde{\hbox{\large$\vartriangleright$}}_{m}&=&P_{m}-\partial^{n}X_{mn}-\frac{1}{4!}\partial^{m_{1}\cdots
m_{4}}X_{mm_{1}\cdots m_{4}}\\\
\tilde{\hbox{\large$\vartriangleright$}}^{m_{1}m_{2}}&=&P^{m_{1}m_{2}}+\partial^{[m_{1}}X^{m_{2}]}-\frac{1}{2}\partial^{m_{1}\cdots
m_{4}}X_{m_{3}m_{4}}\\\ \tilde{\hbox{\large$\vartriangleright$}}^{m_{1}\cdots
m_{5}}&=&P^{m_{1}\cdots m_{5}}-\frac{1}{4!}\partial^{[m_{1}\cdots
m_{4}}X^{m_{5}]}\end{array}\right.~{}~{}~{}.$ (C.16)

The Virasoro constraints are the followings

$\displaystyle\left\\{\begin{array}[]{rcl}{\cal
S}^{m}&=&\frac{1}{2}\hbox{\large$\vartriangleright$}_{n}\hbox{\large$\vartriangleright$}^{nm}=0\\\
{\cal S}^{m_{1}\cdots
m_{4}}&=&\frac{1}{2}\left[\hbox{\large$\vartriangleright$}_{n}\hbox{\large$\vartriangleright$}^{nm_{1}\cdots
m_{4}}+\frac{1}{8}\hbox{\large$\vartriangleright$}^{[m_{1}m_{2}}\hbox{\large$\vartriangleright$}^{m_{3}m_{4}]}\right]=0\\\
{\cal
T}&=&\frac{1}{4}\left[\hbox{\large$\vartriangleright$}_{m}\delta^{mn}\hbox{\large$\vartriangleright$}_{n}+\frac{1}{2}\hbox{\large$\vartriangleright$}^{m_{1}m_{2}}\delta_{m_{1}n_{1}}\delta_{m_{2}n_{2}}\hbox{\large$\vartriangleright$}^{n_{1}n_{2}}\right.\\\
&&\left.+\frac{1}{5!}\hbox{\large$\vartriangleright$}^{m_{1}\cdots
m_{5}}\delta_{m_{1}n_{2}}\cdots\delta_{m_{5}n_{5}}\hbox{\large$\vartriangleright$}^{n_{1}\cdots
n_{5}}\right]=0\end{array}\right.$ (C.20)

For the tensor calculation there are several useful relations to obtain the
Virasoro algebras for the totally antisymmetric tensors $t^{m_{1}m_{2}}$ and
$T^{m_{1}\cdots m_{4}}$ which are obtained by the totally anti-symmetric 6
indices as

$\displaystyle\frac{1}{3!}V^{[m_{1}}\epsilon^{m_{2}m_{3}m_{4}]n_{1}n_{2}}$
$\displaystyle=$
$\displaystyle-V^{[n_{1}}\epsilon^{n_{2}]m_{1}m_{2}m_{3}m_{4}}$
$\displaystyle\frac{1}{4}t^{[m_{1}m_{2}}T^{m_{3}m_{4}]n_{1}n_{2}}$
$\displaystyle=$ $\displaystyle-t^{n_{1}n_{2}}T^{m_{1}\cdots m_{4}}$
$\displaystyle\frac{1}{4}\epsilon^{ml_{1}l_{2}[n_{1}n_{2}}T^{n_{3}n_{4}]}$
$\displaystyle=$ $\displaystyle\epsilon^{mn_{1}\cdots
n_{4}}T^{l_{1}l_{2}}~{}~{}~{}.$ (C.21)

##  Appendix D Double vielbein formulation

The double vielbein formulation in the simplest example is explain in this
appendix. The selfduality constraint in the T-theory Hamiltonian with a flat
worldsheet is the anti-selfduality current is 0: The 2D-dimensional selfdual
current (the covariant derivative) and the 2D-dimensional selfduality current
(the symmetry generator current) are given by

$\displaystyle\left\\{\begin{array}[]{ccl}\hbox{\large$\vartriangleright$}_{M}&=&P_{M}+\partial_{\sigma}X^{N}\eta_{NM}\\\
\tilde{\hbox{\large$\vartriangleright$}}_{M}&=&P_{M}-\partial_{\sigma}X^{N}\eta_{NM}\end{array}\right.~{}~{}~{}.$
(D.3)

When the 2D-dimensional coordinate $X^{M}$ is written in terms of the
D-dimensional coordinates as $(x,y)$ and the canonical conjugates as
$(p_{x},p_{y})$ for the O($D,D$) invariant metric
$\eta_{MN}=\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right)$, the
currents are written as

$\displaystyle\hbox{\large$\vartriangleright$}_{M}=\left\\{\begin{array}[]{ccl}p_{x}+\partial_{\sigma}y\\\
p_{y}+\partial_{\sigma}x\end{array}\right.~{}~{},~{}~{}\tilde{\hbox{\large$\vartriangleright$}}_{M}=\left\\{\begin{array}[]{ccl}p_{x}-\partial_{\sigma}y\\\
p_{y}-\partial_{\sigma}x\end{array}\right.$ $\displaystyle~{}~{}~{}.$ (D.8)

The selfduality constraint is the anti-selfdual current is 0
$\tilde{\hbox{\large$\vartriangleright$}}_{M}=0$ in the usual formulation. By
using the selfduality constraint $p_{y}=\partial_{\sigma}x$, the selfdual
current reduces into the D-dimensional momenta and the winding modes
$\hbox{\large$\vartriangleright$}_{M}\to(p_{x},\partial_{\sigma}x)$.

When the Hamiltonian is made from only the selfdual currents, the Hamiltonian
form Lagrangian gives a chiral scalar Lagrangian where the term
$(\partial_{\sigma}x){}^{2}$ is absent.

$\displaystyle I$ $\displaystyle=$ $\displaystyle\int
L~{}~{},~{}~{}L=\dot{x}p-H~{}~{}$ $\displaystyle H$ $\displaystyle=$
$\displaystyle\frac{g}{2}\hbox{\large$\vartriangleright$}{}^{2}~{}~{}\to
L=\frac{1}{g}(\dot{X}^{2}+2g\dot{X}~{}\partial_{\sigma}X)~{}~{}~{}.$

But adding the squared anti-selfdual current as a constraint with the Lagrange
multiplier $\tilde{g}$ leads to the worldsheet covariant action as

$\displaystyle H$ $\displaystyle=$
$\displaystyle\frac{g}{4}\hbox{\large$\vartriangleright$}{}^{2}+\frac{\tilde{g}}{4}\tilde{\hbox{\large$\vartriangleright$}}{}^{2}\to~{}L=\frac{1}{g+\tilde{g}}(\dot{X}-g\partial_{\sigma}X)(\dot{X}+\tilde{g}\partial_{\sigma}X)~{}~{}~{}.$
(D.9)

This is rewritten in terms of the selfdual and the anti-selfdual currents as

$\displaystyle
L=\displaystyle\frac{1}{2g}J_{\mathrm{SD}}J_{\overline{\mathrm{SD}}}+\frac{g-\tilde{g}}{2g(g+\tilde{g})}J_{\overline{\mathrm{SD}}}{}^{2}$
(D.10)
$\displaystyle\left\\{\begin{array}[]{ccl}J_{\mathrm{SD}}{}^{M}&=&\dot{X}^{M}+g\hat{\eta}^{MN}\eta_{NL}\partial_{\sigma}X^{L}\\\
J_{\overline{\mathrm{SD}}}&=&\dot{X}^{M}-g\hat{\eta}^{MN}\eta_{NL}\partial_{\sigma}X^{L}\end{array}\right.$
(D.13) $\displaystyle
J_{\mathrm{SD}}{}^{M}=\left\\{\begin{array}[]{l}\dot{x}+g\partial_{\sigma}y\\\
\dot{y}+{g}\partial_{\sigma}x\end{array}\right.~{}~{},~{}~{}J_{\overline{\mathrm{SD}}}=\left\\{\begin{array}[]{l}\dot{x}-g\partial_{\sigma}y\\\
\dot{y}-{g}\partial_{\sigma}x\end{array}\right.$ $\displaystyle~{}~{}~{}.$
(D.18)

The first term in the Lagrangian is the free kinetic term while the second
term is the selfduality constraint in a bilinear form. The bilinear form
constraint reduces into the anti-selfdual current to be 0, which relates the
doubled coordinates $x$ and $y$ as the usual selfduality constraint.

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# An elementary proof of Ramanujan’s identity for odd zeta values

Sarth Chavan  Euler Circle, Palo Alto, California 94306.
<EMAIL_ADDRESS>
###### Abstract.

The main goal of this article is to present an elementary proof of Ramanujan’s
identity for odd zeta values. Our proof solely relies on a Mittag-Leffler type
expansion for hyperbolic cotangent function and Euler’s identity for even zeta
values.

###### Key words and phrases:

Riemann zeta function, Ramanujan’s identity for odd zeta values

###### 2020 Mathematics Subject Classification:

Primary 11M06

##  1\. A Brief History and Introduction

The Riemann zeta function $\zeta(s)$ is one of the most important special
functions of Mathematics. While the critical strip $0<\Re\left(s\right)<1$ is
undoubtedly the most important region in the complex plane on account of the
unsolved problem regarding location of non-trivial zeros of $\zeta(s)$,
namely, the Riemann Hypothesis, the right-half plane $\Re\left(s\right)>1$
also has its own share of interesting unsolved problems to contribute to.

It is quite well known that many number theoretic properties of odd zeta
values are nowadays still unsolved mysteries, such as the rationality,
transcendence and existence of closed-forms. Only in 1978 did Apéry [2]
famously proved that $\zeta(3)$ is irrational. This was later reproved in a
variety of ways by several authors, in particular Beukers [10] who devised a
simple approach involving certain integrals over $[0,1]^{3}$.

In the early 2000s, an important work of Rivoal [21], and Ball and Rivoal [4]
determined that infinitely many values of $\zeta$ at odd integers are
irrational, and the work of Zudilin [27] proved that at least one among
$\zeta(5),\zeta(7),\zeta(9)$ and $\zeta(11)$ is irrational. A very recent
result due to Rivoal and Zudilin [22] states that at least two of the numbers
$\zeta(5),\zeta(7),\ldots,\zeta(69)$ are irrational. Moreover, for any pair of
positive integers $a$ and $b$, Haynes and Zudilin [17, Theorem 1] have shown
that either there are infinitely many $m\in\mathbb{N}$ for which $\zeta(am+b)$
is irrational, or the sequence $\\{q_{m}\\}_{m=1}^{\infty}$ of common
denominators of the rational elements of the set
$\\{\zeta(a+b),\ldots,\zeta(am+b)\\}$ grows super-exponentially, that is,
$q_{m}^{1/m}\to\infty$ as $m\to\infty$.

Despite these advances, to this day no value of $\zeta(2n+1)$ with $n\geqslant
2$ is known to be irrational. A folklore conjecture states that the numbers
$\pi$, $\zeta(3),\zeta(5),\zeta(7),\ldots$ are algebraically independent over
the rationals. This conjecture is predicted by the Grothendieck’s period
conjecture for mixed Tate motives. But both conjectures are far out of reach
and we do not even know the transcendence of a single odd zeta value.

One should mention that Brown [11] has in the past few years outlined a simple
geometric approach to understand the structures involved in Beukers’s proof of
irrationality of $\zeta(3)$ and how this may generalize to other odd zeta
values.

Ramanujan made many beautiful and elegant discoveries in his short life of 32
years. One of the most remarkable formulas suggested by Ramanujan that has
attracted the attention of several mathematicians over the years is the
following intriguing identity involving the odd values of the Riemann zeta
function [6, 1.2]:

######  Theorem 1.1 (Ramanujan’s Formula for $\zeta(2m+1)$).

If $\alpha$ and $\beta$ are positive numbers such that $\alpha\beta=\pi^{2}$
and if $m$ is a positive integer, then we have

$\displaystyle\alpha^{-m}\,\left\\{\dfrac{1}{2}\,\zeta(2m+1)+\sum_{n=1}^{\infty}\dfrac{n^{-2m-1}}{e^{2\alpha
n}-1}\right\\}$
$\displaystyle-(-\beta)^{-m}\,\left\\{\dfrac{1}{2}\,\zeta(2m+1)+\sum_{n=1}^{\infty}\dfrac{n^{-2m-1}}{e^{2\beta
n}-1}\right\\}$ (1.1)
$\displaystyle=2^{2m}\sum_{k=0}^{m+1}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{2m-2k+2}}{\left(2k\right)!\left(2m-2k+2\right)!}\,\alpha^{m-k+1}\beta^{k}.$

where $B_{m}$ denotes the $m$-th Bernoulli number with the exponential
generating function

$\sum_{m\geqslant 0}B_{m}\frac{z^{m}}{m!}=\dfrac{z}{\exp\left(z\right)-1}.$

Theorem 1.1 appears as Entry 21 in Chapter 14 of Ramanujan’s second notebook
[5, 173]. The first published proof of Theorem 1.1 is due to S.L. Marulkar
[20] although he was not aware that this formula can be found in Ramanujan’s
Notebooks. Grosswald too rediscovered this formula and studied it more
generally in [15], [14]. Berndt [8, Theorem 2.2] derived a general formula
from which both Euler’s identity for even zeta values (1.1) and Ramanujan’s
identity for odd zeta values follow as special cases, thus showing that
Euler’s and Ramanujan’s formulas are natural companions of each other.

Ramanujan’s identity has a number of applications. For example, a contemporary
interpretation of the above identity, as given for instance in [16], is that
it encodes fundamental transformation properties of Eisenstein series on the
full modular group and their Eichler integrals. This observation is extended
in [9, Section 5] to weight $2k+1$ Eisenstein series of level $2$ through
secant Dirichlet series. Moreover, Ramanujan’s identity also has applications
in theoretical computer science [18] in the analysis of special data
structures and algorithms. More specifically, it is used there to achieve
certain distribution results on random variables related to data structures
called _tries_.
B. Berndt and A. Straub in their article [6, 3.1] show that identity (1.1) is
equivalent to

$\displaystyle\alpha^{-m}\sum_{n=1}^{\infty}\dfrac{\coth\left(\alpha
n\right)}{n^{2m+1}}$
$\displaystyle-(-\beta)^{-m}\sum_{n=1}^{\infty}\dfrac{\coth\left(\beta
n\right)}{n^{2m+1}}$ (1.2)
$\displaystyle=-2^{2m+1}\sum_{k=0}^{m+1}\dfrac{\left(-1\right)^{k}B_{2k}\,B_{2m+2-2k}}{\left(2k\right)!\left(2m+2-2k\right)!}\,\alpha^{m+1-k}\beta^{k}.$

Letting $\alpha=\beta=\pi$ and replacing $m$ with $2m+1$, we get the following
[6, 3.2]:

######  Proposition 1.2.

The following identity holds

(1.3) $\sum_{n=1}^{\infty}\dfrac{\coth\left(\pi
n\right)}{n^{4m+3}}=2^{4m+2}\pi^{4m+3}\sum_{k=0}^{2m+2}\dfrac{\left(-1\right)^{k+1}B_{2k}\,B_{4m+4-2k}}{\left(2k\right)!\left(4m+4-2k\right)!}$

where as before $m\in\mathbb{N}$ and $B_{m}$ denotes the $m$-th Bernoulli
number.

This variation was apparently first established by M. Lerch [19]. Later proofs
were given by G.N. Watson [26], H.F. Sandham [23], J.R. Smart [25] and F.P.
Sayer [24].

The main purpose of this article is to present an elementary proof of Theorem
1.1 and Proposition 1.2. Our proof solely relies on a Mittag-Leffler type
expansion for hyperbolic cotangent function and Euler’s identity for even zeta
values

##  2\. Preliminaries

######  Proposition 2.1 (Euler’s Formula for $\zeta(2m)$).

For $m\geqslant 1$, we have

(2.1)
$\zeta(2m)=\left(2\pi\right)^{2m}\dfrac{\left(-1\right)^{m-1}B_{2m}}{2\left(2m\right)!}.$

Euler’s identity for even zeta values (2.1) not only provides an elegant
formula for evaluating $\zeta(2m)$ for any integer $m\geqslant 1$, but it also
gives us information about the arithmetical nature of even zeta values. For a
fascinating account of the history and proof of this formula, we refer the
reader to [3].

######  Proposition 2.2 (Mittag-Leffler expansion for hyperbolic cotangent
function).

Let $x\in\mathbb{C}\setminus{\\{0\\}}$, then the following identity holds

(2.2) $\coth\left(\pi x\right)=\dfrac{1}{\pi
x}+\dfrac{2x}{\pi}\sum_{k=1}^{\infty}\dfrac{1}{x^{2}+k^{2}}.$

Note that equation (2.2) appears as identity 1.421.4 in [13].

##  3\. Proof of Proposition 1.2

Replacing $x$ with $n$, dividing both sides of identity $(\ref{ML2})$ by
$n^{4m+2}$ and summing over $n$ produces

$\dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{\pi\coth(\pi
n)}{n^{4m+3}}-\dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m+4}}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\dfrac{1}{n^{4m+2}\,(k^{2}+n^{2})}.$

Since the above double sum is absolutely convergent, we may switch the order
of summation to get

$\dfrac{1}{2}\left(\sum_{n=1}^{\infty}\dfrac{\pi\coth(\pi
n)}{n^{4m+3}}-\sum_{n=1}^{\infty}\dfrac{1}{n^{4m+4}}\right)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m+2}\,(k^{2}+n^{2})}.$

Next, we simply notice that

$\dfrac{1}{n^{4m+2}\left(k^{2}+n^{2}\right)}=\dfrac{1}{n^{4m}k^{2}}\left(\dfrac{1}{n^{2}}-\dfrac{1}{k^{2}+n^{2}}\right).$

Thus, we have

$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\dfrac{1}{n^{4m+2}\left(k^{2}+n^{2}\right)}=\zeta(2)\,\zeta(4m+2)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m}k^{2}\left(k^{2}+n^{2}\right)}.$

Similarly, we find that

$\dfrac{1}{n^{4m}k^{2}\,(k^{2}+n^{2})}=\dfrac{n^{2}}{n^{4m}k^{4}}\left(\dfrac{1}{n^{2}}-\dfrac{1}{k^{2}+n^{2}}\right).$

Thus, we have

$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m}k^{2}\left(k^{2}+n^{2}\right)}=\zeta(4)\,\zeta(4m)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m-2}k^{4}\left(k^{2}+n^{2}\right)}.$

This recursive pattern that produces even zeta values continues as follows:

$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m-2}k^{4}\left(k^{2}+n^{2}\right)}=\zeta(6)\,\zeta(4m-2)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m-4}k^{6}\left(k^{2}+n^{2}\right)}$
$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m-4}k^{6}\left(k^{2}+n^{2}\right)}=\zeta(8)\,\zeta(4m-4)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m-6}k^{8}\left(k^{2}+n^{2}\right)}$

and so on. At the end, we get

$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{2}k^{4m}\left(k^{2}+n^{2}\right)}=\zeta(4m+2)\,\zeta(2)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{k^{4m+2}\left(k^{2}+n^{2}\right)}.$

Therefore, we obtain the final result

$\displaystyle\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m+2}\left(k^{2}+n^{2}\right)}$
$\displaystyle=\zeta(2)\,\zeta(4m+2)-\zeta(4)\,\zeta(4m)+\cdots+\zeta(4m+2)\,\zeta(2)$
(3.1)
$\displaystyle-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{k^{4m+2}\left(k^{2}+n^{2}\right)}$

which can be easily proved by induction on $m$. Next, we observe that, by
symmetry,

$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m+2}\left(k^{2}+n^{2}\right)}=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{k^{4m+2}\left(k^{2}+n^{2}\right)}.$

Therefore we have

$\displaystyle
2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{n^{4m+2}\left(k^{2}+n^{2}\right)}$
$\displaystyle=\left(-1\right)^{m}\zeta(2m+2)\,\zeta(2m+2)$
$\displaystyle+2\sum_{k=0}^{m-1}\left(-1\right)^{k}\zeta(4m+2-2k)\,\zeta(2k+2).$

Putting all things together produces

$\displaystyle\sum_{n=1}^{\infty}\dfrac{\pi\coth(\pi n)}{n^{4m+3}}$
$\displaystyle=\zeta(4m+4)+(-1)^{m}\zeta(2m+2)\,\zeta(2m+2)$
$\displaystyle+2\sum_{k=1}^{m}(-1)^{k-1}\,\zeta(4m-2k+4)\,\zeta(2k)$

where $k$ has been replaced with $k-1$ in the summand. Thus, it suffices to
prove that

$\displaystyle\zeta(4m+4)+\left(-1\right)^{m}$
$\displaystyle\zeta(2m+2)\,\zeta(2m+2)+2\sum_{k=1}^{m}\left(-1\right)^{k-1}\zeta(4m-2k+4)\,\zeta(2k)$
(3.2) $\displaystyle=\,$
$\displaystyle\,2^{4m+2}\pi^{4m+4}\sum_{k=0}^{2m+2}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{4m+4-2k}}{\left(2k\right)!\left(4m+4-2k\right)!}.$

After replacing $m$ with $m-1$ in the above equation, it suffices to prove
that

$\displaystyle\zeta(4m)+\left(-1\right)^{m-1}$
$\displaystyle\zeta(2m)\,\zeta(2m)+2\sum_{k=1}^{m-1}\left(-1\right)^{k-1}\zeta(4m-2k)\,\zeta(2k)$
(3.3) $\displaystyle=\,$
$\displaystyle\,2^{4m-2}\pi^{4m}\sum_{k=0}^{2m}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{4m-2k}}{\left(2k\right)!\left(4m-2k\right)!}$

which is indeed just a matter of substitution; using Euler’s identity for
$\zeta(2m)$, we have

$\displaystyle\zeta(4m)+\left(-1\right)^{m-1}\zeta(2m)\,\zeta(2m)+2\sum_{k=1}^{m-1}(-1)^{k-1}\,\zeta(4m-2k)\,\zeta(2k)$
(3.4)
$\displaystyle=2^{4m-2}\pi^{4m}\left\\{-\dfrac{B_{4m}}{\left(4m\right)!}+\dfrac{\left(-1\right)^{m-1}B_{2m}\,B_{2m}}{\left(2m\right)!\left(2m\right)!}+2\sum_{k=1}^{m-1}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{4m-2k}}{\left(2k\right)!\left(4m-2k\right)!}\right\\}$

The summand in equation (3.3) is invariant under the substitution $k\mapsto
2m-k$, so if we split it according to
$k\in\\{0,2m\\},\,k\in\\{1,2,\ldots,m+1,\ldots,2m-1\\}$ and $k=m$, we get

(3.5)
$\sum_{k=0}^{2m}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{4m-2k}}{\left(2k\right)!\left(4m-2k\right)!}=-\dfrac{B_{4m}}{\left(4m\right)!}+\dfrac{\left(-1\right)^{m-1}B_{2m}\,B_{2m}}{\left(2m\right)!\left(2m\right)!}+2\sum_{k=1}^{m-1}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{4m-2k}}{\left(2k\right)!\left(4m-2k\right)!}.$

Combining equations (3.3), (3.4) and (3.5) gives us the desired result.
$\blacksquare$

##  4\. Proof of Theorem 1.1

Substituting $x\mapsto\left(\alpha k/\pi\right)$, with
$\alpha\in\mathbb{R}_{>0}$, in identity (2.2) yields

(4.1) $\dfrac{1}{e^{2\alpha k}-1}=-\dfrac{1}{2}+\dfrac{1}{2\alpha
k}+\sum_{m=1}^{\infty}\dfrac{\alpha k}{\pi^{2}m^{2}+\alpha^{2}k^{2}}.$

Dividing both sides of the above identity by $\alpha^{n}k^{2n+1}$ and summing
over $k$ produces

$\alpha^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\alpha
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}=\dfrac{\zeta(2n+2)}{2\alpha^{n+1}}+\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{\alpha^{n+1}}{\left(\alpha^{2}k^{2}\right)^{n}\left(\pi^{2}m^{2}+\alpha^{2}k^{2}\right)}.$

Next, we notice that, for an arbitrary positive integer $n$ and two arbitrary
non-vanishing sequences $\\{x_{k}\\}$ and $\\{z_{m}\\}$, we have

(4.2)
$\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{x_{k}^{n}\left(x_{k}+z_{m}\right)}=\sum_{k=1}^{\infty}\frac{1}{x_{k}^{n}}\sum_{m=1}^{\infty}\frac{1}{z_{m}}-\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{x_{k}^{n-1}z_{m}\left(x_{k}+z_{m}\right)}$

assuming that the sums involved are convergent. Using the same recursive
technique as in the previous section, we find that

$\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{x_{k}^{n}\left(x_{k}+z_{m}\right)}=\sum_{k=1}^{\infty}\frac{1}{x_{k}^{n}}\sum_{m=1}^{\infty}\frac{1}{z_{m}}-\sum_{k=1}^{\infty}\frac{1}{x_{k}^{n-1}}\sum_{m=1}^{\infty}\frac{1}{z_{m}^{2}}$
(4.3)
$+\sum_{k=1}^{\infty}\frac{1}{x_{k}^{n-2}}\sum_{m=1}^{\infty}\frac{1}{z_{m}^{3}}+\cdots+(-1)^{n-1}\sum_{k=1}^{\infty}\frac{1}{x_{k}}\sum_{m=1}^{\infty}\frac{1}{z_{m}^{n}}+(-1)^{n}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{z_{m}^{n}\left(x_{k}+z_{m}\right)}.$

For notational convenience, set

$\sum_{k=1}^{\infty}\frac{1}{x_{k}^{n}}=\zeta_{x}(n),\quad\sum_{k=1}^{\infty}\frac{1}{z_{m}^{n}}=\zeta_{z}(n).$

These are also known as _quasisymmetric zeta function_ (the correct analog of
a Riemann zeta function in the ring of quasisymmetric functions).

Thus, equation (4.3) can now be rewritten as

(4.4)
$\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{x_{k}^{n}\left(x_{k}+z_{m}\right)}=\sum_{p=0}^{n-1}(-1)^{p}\zeta_{x}(n-p)\,\zeta_{z}(p+1)+(-1)^{n}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{z_{m}^{n}\left(x_{k}+z_{m}\right)}$

Substituting $x_{k}\mapsto\alpha^{2}k^{2}$ and $z_{m}\mapsto\pi^{2}m^{2}$ in
identity (4.4) produces

$\displaystyle\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{\left(\alpha^{2}k^{2}\right)^{n}\left(\alpha^{2}k^{2}+\pi^{2}m^{2}\right)}$
$\displaystyle=\sum_{p=0}^{n-1}\left(-1\right)^{p}\dfrac{\zeta(2n-2p)}{\alpha^{2n-2p}}\,\dfrac{\zeta(2p+2)}{\pi^{2p+2}}$
(4.5)
$\displaystyle+\left(-1\right)^{n}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{\left(\pi^{2}m^{2}\right)^{n}\left(\alpha^{2}k^{2}+\pi^{2}m^{2}\right)}.$

Euler’s formula for $\zeta(2m)$ allows us to transform the zetas into
Bernoullis as follows:

$\sum_{p=0}^{n-1}\left(-1\right)^{p}\dfrac{\zeta(2n-2p)}{\alpha^{2n-2p}}\dfrac{\zeta(2p+2)}{\pi^{2p+2}}=\left(-1\right)^{n}2^{2n}\sum_{p=0}^{n-1}\left(\dfrac{\pi}{\alpha}\right)^{2n-2p}\dfrac{\left(-1\right)^{p-1}B_{2n-2p}\,B_{2p+2}}{\left(2n-2p\right)!\left(2p+2\right)!}.$

Moreover, using $\alpha\beta=\pi^{2}$ and replacing $p$ with $p-1$ in the
summand produces

$\sum_{p=0}^{n-1}\left(-1\right)^{p}\,\dfrac{\zeta(2n-2p)}{\alpha^{2n-2p}}\dfrac{\zeta(2p+2)}{\pi^{2p+2}}=\left(-1\right)^{n}2^{2n}\sum_{p=1}^{n}\left(\dfrac{\beta}{\alpha}\right)^{n-p+1}\dfrac{\left(-1\right)^{p}B_{2n-2p+2}\,B_{2p}}{\left(2n-2p+2\right)!\left(2p\right)!}.$

Substituting this expression in equation (4.5) we get

$\displaystyle\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{\left(\alpha^{2}k^{2}\right)^{n}\left(\alpha^{2}k^{2}+\pi^{2}m^{2}\right)}$
$\displaystyle=\left(-1\right)^{n}2^{2n}\sum_{p=1}^{n}\left(\dfrac{\beta}{\alpha}\right)^{n-p+1}\dfrac{\left(-1\right)^{p}B_{2n-2p+2}\,B_{2p}}{\left(2n-2p+2\right)!\left(2p\right)!}$
(4.6)
$\displaystyle+\left(-1\right)^{n}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{\left(\pi^{2}m^{2}\right)^{n}\left(\alpha^{2}k^{2}+\pi^{2}m^{2}\right)}$

Putting all things together produces

$\alpha^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\alpha
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}=\left(-1\right)^{n}\alpha^{n+1}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{(\pi^{2}m^{2})^{n}(\alpha^{2}k^{2}+\pi^{2}m^{2})}$
$+\,\dfrac{\zeta(2n+2)}{2\alpha^{n+1}}+\alpha^{n+1}\left\\{=\left(-1\right)^{n}2^{2n}\sum_{p=1}^{n}\left(\dfrac{\beta}{\alpha}\right)^{n-p+1}\dfrac{\left(-1\right)^{p}B_{2n-2p+2}\,B_{2p}}{\left(2n-2p+2\right)!\left(2p\right)!}\right\\}.$

Next, we notice that the remaining zeta term

$\dfrac{\zeta(2n+2)}{2\alpha^{n+1}}=\dfrac{\left(-1\right)^{n}}{2\alpha^{n+1}}\dfrac{\left(2\pi\right)^{2n+2}B_{2n+2}}{2\left(2n+2\right)!}=\dfrac{\left(-1\right)^{n}\beta^{n+1}2^{2n}B_{2n+2}}{\left(2n+2\right)!}$

is the missing $p=0$-th term in the summand. Thus, we have

$\displaystyle\alpha^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\alpha
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}$
$\displaystyle-\alpha^{n+1}\left\\{\dfrac{\left(-1\right)^{n}}{\pi^{2n}}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{m^{2n}\left(\alpha^{2}k^{2}+m^{2}\pi^{2}\right)}\right\\}$
(4.7)
$\displaystyle=\left(-1\right)^{n}2^{2n}\sum_{p=0}^{n}\dfrac{\left(-1\right)^{p}B_{2n-2p+2}\,B_{2p}}{\left(2n-2p+2\right)!\left(2p\right)!}\,\alpha^{p}\beta^{n-p+1}.$

Notice that, using $\alpha\beta=\pi^{2}$ we obtain

$\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{m^{2n}\left(\alpha^{2}k^{2}+m^{2}\pi^{2}\right)}=\dfrac{1}{\alpha}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{m^{2n}\left(\alpha
k^{2}+\beta
m^{2}\right)}=\dfrac{1}{\alpha}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{k^{2n}\left(\alpha^{2}m^{2}+\beta
k^{2}\right)},$

where the last equality is obtained by replacing $k$ by $m$ and $m$ by $k$ so
that we get

(4.8)
$\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{m^{2n}\left(\alpha^{2}k^{2}+m^{2}\pi^{2}\right)}=\dfrac{\beta}{\alpha}\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{k^{2n}\left(\beta^{2}k^{2}+m^{2}\pi^{2}\right)}$

Dividing both sides of identity (2.2) by $k^{2n+1}$ and summing over $k$
produces

(4.9)
$\beta\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{k^{2n}\left(\beta^{2}k^{2}+m^{2}\pi^{2}\right)}=\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\beta
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)-\dfrac{1}{2\beta}\,\zeta(2n+2)$

where we have replaced $\alpha$ with $\beta$. Thus, we have

$\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{m^{2n}\left(\alpha^{2}k^{2}+m^{2}\pi^{2}\right)}=\dfrac{1}{\alpha}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\beta
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)-\dfrac{1}{2\beta}\,\zeta(2n+2)\right\\}.$

Finally, substituting this expression in equation (4.7) yields

$\displaystyle\alpha^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\alpha
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}-\left(-1\right)^{n}2^{2n}\sum_{p=0}^{n}\dfrac{\left(-1\right)^{p}B_{2n-2p+2}\,B_{2p}}{\left(2n-2p+2\right)!\left(2p\right)!}\,\alpha^{p}\,\beta^{n-p+1}$
$\displaystyle=\dfrac{\left(-1\right)^{n}\alpha^{n}}{\pi^{2n}}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\beta
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}-\dfrac{\left(-1\right)^{n}\alpha^{n+1}}{\pi^{2n}}\left\\{\dfrac{1}{2\alpha\beta}\,\zeta(2n+2)\right\\}$
$=(-\beta)^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\beta
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}-\dfrac{\left(-1\right)^{n}\alpha^{n+1}}{\pi^{2n+2}}\dfrac{\left(-1\right)^{n}2^{2n+2}\pi^{2n+2}B_{2n+2}}{4\left(2n+2\right)!}$

Next, we notice that the remaining term:

$-\dfrac{\left(-1\right)^{n}\alpha^{n+1}}{\pi^{2n+2}}\dfrac{\left(-1\right)^{n}2^{2n+2}\pi^{2n+2}B_{2n+2}}{4\left(2n+2\right)!}=\dfrac{\left(-1\right)^{2n+1}2^{2n}B_{2n+2}}{\left(2n+2\right)!}\,\alpha^{n+1}$

is the $p=n+1$-st term in the summand. Therefore, we deduce that

(4.10)
$\displaystyle\alpha^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\alpha
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}$
$\displaystyle-\left(-\beta\right)^{-n}\left\\{\sum_{k=1}^{\infty}\dfrac{k^{-2n-1}}{e^{2\beta
k}-1}+\dfrac{1}{2}\,\zeta(2n+1)\right\\}$
$\displaystyle=\left(-1\right)^{n}2^{2n}\sum_{p=0}^{n+1}\dfrac{\left(-1\right)^{p}B_{2n-2p+2}\,B_{2p}}{\left(2n-2p+2\right)!\left(2p\right)!}\,\alpha^{p}\beta^{n-p+1}.$

Replacing $p$ with $n-p+1$ in the above identity gives us the desired result.
$\blacksquare$

##  5\. Acknowledgements

The author would like to thank Christophe Vignat for his guidance and support
throughout the completion of this work. The author is very much thankful to
the reviewer for his/her valuable comments towards the improvement of the
manuscript. The author would also like to thank Bruce Berndt and Atul Dixit
for endorsing him on arXiv.

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