Ramanujan’s formula for odd zeta values: a proof by Mittag-Leffler expansion and applications

Abstract

Using Mittag-Leffler expansion a novel proof of Ramanujan’s famous formula for $\zeta (2m-1)$ is presented. Here, the formula can be derived by Taylor series of a generating function and its Mittag-Leffler expansion. Furthermore, generalized formulas for the fast-converging series of Plouffe are offered. But Ramanujan’s equation not only produces series for odd zeta values, surprisingly it is shown that in the limit to its singularity it provides the identities for even zeta values, namely Euler’s classical formula. Finally, a new triangle identity of Ramanujan’s formula is presented, hinting that properties and symmetries of the equation are far from all uncovered. For this work, a new representation of the equations is introduced, that significantly simplifies analyses and applications.

1 A proof by Mittag-Leffler expansion

Whereas Euler’s well-known formula provides a calculation of zeta values of even integers, $\zeta (2m)$, no closed-form solutions are known for zeta values of odd integers, $\zeta (2m-1)$. Lerch [6] and Ramanujan [1, 11] presented impressive formulas, which give rapidly converging series for $\zeta (2m-1)$, for a comprehensive overview see also Berndt and Straub [2]. Here, Lerch’s formula is the special case $a=b=\pi$ of Ramanujan’s more general expression: For $ab=\pi ^2$,

$$\begin{aligned} {} -2^{2m+1} \sum _{j=0}^{m+1} \,(-1)^j&a^{m+1-j} b^j\hspace{0.55542pt}\frac{B_{2j} B_{2m+2-2j}}{(2j)!\,(2m+2-2j)!} \\&= a^{-m} \sum _{n=1}^\infty \, \frac{\coth \hspace{0.55542pt}(an)}{n^{2m+1}} -(-b)^{-m} \sum _{n=1}^\infty \, \frac{\coth \hspace{0.55542pt}(bn)}{n^{2m+1}} \\&= a^{-m} \biggl [\zeta (2m+1) + \sum _{n=1}^\infty \,\frac{2}{n^{2m+1}(e^{2na}-1)}\biggr ]\\&\qquad \quad - (-b)^{-m} \biggl [\zeta (2m+1) + \sum _{n=1}^\infty \,\frac{2}{n^{2m+1}(e^{2nb}-1)}\biggr ] . \end{aligned}$$

Proofs of this formula are the subject of many publications, mostly using the partial fraction decomposition of $\coth$ function, e.g. [3, 8], or residue calculus, e.g. [7, 13].

In this work, a novel approach via Mittag-Leffler expansion of the generating function is presented. Without loss of generality, we choose $a=\pi /\alpha$ and $b=\pi \alpha$, which simplifies the treatment hereinafter. Furthermore, by replacing $m\rightarrow m-1$ and multiplying the equation by $(2\pi )^{-m}$, Ramanujan’s formula can be expressed as follows:

Theorem 1.1

For $m\geqslant 2$,

$$\begin{aligned} \begin{aligned} F_m(\alpha )&= -\sum _{j=0}^m \,(-1)^j \alpha ^{2j-m} \widetilde{B}_{2j}\widetilde{B}_{2m-2j} \\&= \alpha ^{m-1} \sum _{n=1}^\infty \, \frac{\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})}{(2\pi n)^{2m-1}} - (-\alpha )^{1-m} \sum _{n=1}^\infty \, \frac{\coth \hspace{0.55542pt}(\pi n\alpha )}{(2\pi n)^{2m-1}} \hspace{0.55542pt}, \end{aligned} \end{aligned}$$

(1)

where, to simplify the notation, $\widetilde{B}_{2k}$ are defined by even Bernoulli numbers $B_{2k}$ as $\widetilde{B}_{2k}=B_{2k}/(2k)!$ and can be calculated recursively,

$$\begin{aligned} \widetilde{B}_0=1, \quad \widetilde{B}_2=\frac{1}{12}\hspace{0.55542pt}, \quad (2k+1)\hspace{0.55542pt}\widetilde{B}_{2k} = -\sum _{j=1}^{k-1} \widetilde{B}_{2j}\widetilde{B}_{2k-2j}, \;\; k\geqslant 2. \end{aligned}$$

(2)

Proof

Ramanujan [2], Sitaramachandrarao [12], and Xu [14] analyzed the partial fraction decomposition of the function $\cot \hspace{0.55542pt}\bigl (\frac{\sqrt{\alpha }x}{2}\bigr )\coth \hspace{0.55542pt}\bigl (\frac{x}{2\sqrt{\alpha }}\bigr )$. As discussed by Berndt and Straub [2], here a Mittag-Leffler expansion would fail, which is due to the occurrence of divergent sums. To avoid this, we include a factor 1/x and carry out the Mittag-Leffler expansion on the function

$$\begin{aligned} \tilde{f}_\alpha (x) = \frac{1}{x}\cot \hspace{0.55542pt}\biggl (\frac{\sqrt{\alpha }x}{2}\biggr )\coth \hspace{0.55542pt}\biggl (\frac{x}{2\sqrt{\alpha }}\biggr ). \end{aligned}$$

The function $\tilde{f}_\alpha (x)$ has a pole of order 3 in $x_0=0$ and poles of order 1 at the real axis in and at the imaginary axis in $x_\textrm{In}=2\pi ni\sqrt{\alpha }$, both for $n\ne 0$. Its principal parts $p_j(x)$, consisting of the negative powers of the corresponding Laurent series $\tilde{f}_{j\alpha }(x)= \sum _{k=-\infty }^{+\infty } c_{jk} (x-x_j)^{k}$, are calculated as $p_j(x)=\lim _{x\rightarrow x_j}\tilde{f}_\alpha (x)$, since $c_{j0}=0$ holds for all poles $x_j$ of $\tilde{f}_\alpha (x)$:

Applying Mittag-Leffler’s theorem, see e.g. [5, pp. 383–386], $\tilde{f}_\alpha (x)$ can be approximated by the sum of its principal parts,

$$\begin{aligned} \tilde{f}_\alpha (x)&= p_0(x) + \sum _{n\ne 0} p_{\,\mathrm Rn}(x) + \sum _{n\ne 0} p_{\, \mathrm In}(x) \\&= \frac{4}{x^3} - \frac{\alpha -\alpha ^{-1}}{3x} + \sum _{n=1}^\infty \frac{2x \coth \hspace{0.55542pt}(\pi n\alpha ^{-1})}{\pi n (x^2 - \alpha ^{-1}(2\pi n)^2)} - \sum _{n=1}^\infty \frac{2x \coth \hspace{0.55542pt}(\pi n\alpha )}{\pi n (x^2 + \alpha (2\pi n)^2)}\hspace{0.55542pt}, \end{aligned}$$

where the principal parts ($\propto n^{-2}$) yield a normal convergence of $\tilde{f}_\alpha (x)$ on ${\textbf{C}}\backslash \{x_j\}$.

For further analysis, we use $f_\alpha (x) = \frac{1}{4}x^3[\tilde{f}_\alpha (x)-p_0(x)]$, which yields

$$\begin{aligned} \begin{aligned} f_\alpha (x)&= \frac{x^2}{4}\, \biggl [ \cot \hspace{0.55542pt}\biggl (\frac{\sqrt{\alpha }x}{2}\biggr ) \coth \hspace{0.55542pt}\biggl (\frac{x}{2\sqrt{\alpha }}\biggr ) + \frac{\alpha -\alpha ^{-1}}{3} \biggr ] - 1 \\&= \sum _{n=1}^\infty \, \frac{x^4}{2\pi n} \,\biggl [ \frac{\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})}{x^2-\alpha ^{-1}(2\pi n)^2} - \frac{\coth \hspace{0.55542pt}(\pi n\alpha )}{x^2+\alpha (2\pi n)^2} \biggr ] . \end{aligned} \end{aligned}$$

(3)

With Taylor series

$$\begin{aligned} \begin{aligned} \frac{x}{2\sqrt{\alpha }} \coth \hspace{0.55542pt}\biggl (\frac{x}{2\sqrt{\alpha }}\biggr )&= \sum _{m=0}^\infty \,\alpha ^{-m} \widetilde{B}_{2m} x^{2m} , \\ \frac{\sqrt{\alpha }x}{2} \cot \hspace{0.55542pt}\biggl (\frac{\sqrt{\alpha }x}{2}\biggr )&= \sum _{m=0}^\infty \,(-\alpha )^m \widetilde{B}_{2m} x^{2m} , \\ \frac{(2\pi n)^{-1}x^4}{x^2-\alpha ^{-1}(2\pi n)^2}&= -\sum _{m=2}^\infty \, \frac{\alpha ^{m-1} x^{2m}}{(2\pi n)^{2m-1}}\hspace{0.55542pt},\\ \frac{(2\pi n)^{-1}x^4}{x^2+\alpha (2\pi n)^2}&= -\sum _{m=2}^\infty \,\frac{(-\alpha )^{1-m} x^{2m}}{(2\pi n)^{2m-1}} \hspace{0.55542pt}, \end{aligned} \end{aligned}$$

where $|x| < 2\pi \min \hspace{0.55542pt}(|\alpha ^{1/2}|,|\alpha ^{-1/2}|)$, and with $\widetilde{B}_0=1$ and $\widetilde{B}_2=1/12$, equation (2), we get the series expansion for $f_\alpha (x)$ as the generating function of $F_m(\alpha )$ for $m\geqslant 2$:

$$\begin{aligned} \begin{aligned} f_\alpha (x)&= - \sum _{m=2}^\infty F_m(\alpha )\hspace{1.111pt}x^{2m} \\&= \sum _{m=2}^\infty x^{2m} \sum _{j=0}^m \,(-1)^j \alpha ^{2j-m} \widetilde{B}_{2j}\widetilde{B}_{2m-2j} \\&= - \sum _{m=2}^\infty x^{2m}\hspace{1.111pt}\biggl [ \alpha ^{m-1} \sum _{n=1}^\infty \frac{\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})}{(2\pi n)^{2m-1}} - (-\alpha )^{1-m} \sum _{n=1}^\infty \frac{\coth \hspace{0.55542pt}(\pi n\alpha )}{(2\pi n)^{2m-1}} \biggr ] , \end{aligned} \end{aligned}$$

(4)

which gives (1) by terms of $x^{2m}$. $\square$

The functions $F_m(\alpha )$ of (1), being calculated by the modified Bernoulli numbers $\widetilde{B}_{2j}$ (where $0\leqslant j \leqslant m$), consist of a linear combination of the two sums $C_m(\alpha ^{-1})$ and $C_m(\alpha )$. Using the identity, $\coth \hspace{0.55542pt}(x)=1 + 2/(\exp \hspace{0.55542pt}(2x)-1)$, the sums $C_m$ can be decomposed into the odd zeta value $\zeta (2\,m-1)=\sum _{n=1}^\infty n^{1-2m}$ and an infinite sum $E_m$. For $m\geqslant 2$ we get

$$\begin{aligned} \begin{aligned} F_m(\alpha )&= -\sum _{j=0}^m\, (- 1)^j \alpha ^{2j-m} \widetilde{B}_{2j}\widetilde{B}_{2m-2j}\\&= \alpha ^{m-1} C_m(\alpha ^{-1}) - (- \alpha )^{1-m} C_m(\alpha ), \\ C_m(\alpha )&= \sum _{n=1}^\infty \,\frac{\coth \hspace{0.55542pt}(\pi n\alpha )}{(2\pi n)^{2m-1}}= \frac{1}{(2\pi )^{2m-1}}\, \bigl [\zeta (2m-1)+E_m(\alpha )\bigr ], \\ E_m(\alpha )&= \sum _{n=1}^\infty \,\frac{2}{n^{2m-1}\, (e^{2\pi n\alpha }-1)} \hspace{0.55542pt}. \end{aligned} \end{aligned}$$

(5)

Thus, the values $\zeta (2m-1)$ can be calculated by the function values $F_m(\alpha )$ and the sums $E_m(\alpha ^{-1})$ and $E_m(\alpha )$, which converge rapidly for any $\textrm{Re}\,\alpha >0$:

$$\begin{aligned} \begin{aligned} \bigl [ \alpha ^{m-1}&- (-\alpha )^{1-m} \bigr ]\hspace{1.111pt}\zeta (2m-1) \\&= {}-(2\pi )^{2m-1} F_m(\alpha ) - \alpha ^{m-1}E_m(\alpha ^{-1}) + (-\alpha )^{1-m} E_m(\alpha ) . \end{aligned} \end{aligned}$$

(6)

This is a reformulation of the equations of Ramanujan [2] and in the case of $\alpha =1$ of Lerch [6].

Looking at (1), the symmetry of the functions $F_m$ is given by

$$\begin{aligned} F_m(\alpha ) = (- 1)^m F_m(- \alpha ) = (- 1)^m F_m(\alpha ^{-1}) = F_m(- \alpha ^{-1}). \end{aligned}$$

Then, $f_\alpha (x)=-\sum _{m=2}^\infty F_m(\alpha )\hspace{0.55542pt}x^{2m}$ applies to $f_\alpha (\pm x) = f_{-\alpha }(\pm ix) = f_{-1/\alpha }(\pm x)= f_{1/\alpha }(\pm ix)$. With (1), $F_m(\alpha )$ has a pole of order m in $\alpha =0$, whereas the singularities of the right-hand expression in $\alpha =ki$ and $\alpha ^{-1}=ki$, $k\in {\textbf{Z}}\backslash \{0\}$, generated by the terms $\coth \hspace{0.55542pt}(\pi n\alpha )$ and $\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})$, are removable.

2 Fast converging series for $\zeta (2m-1)$

In particular, the case $\alpha =1$ is the subject of various publications, see e.g. Plouffe [9, 10], Ghusayni [4], and Vepstas [13]. Furthermore, fast converging series of $\zeta (2m-1)$ for $\alpha =2$ and $\alpha =1+i$ are used by Vepstas [13] and Plouffe [9, 10]. Based on that work, using (5) and (6) we derive a generalized set of formulas for fast converging series of $\zeta (2m-1)$ and present the values of the associated coefficients.

(a) For $\alpha =1$ it is $C_m(\alpha ^{-1})=C_m(\alpha )=C(1)$, see (5). Here we have $F_{2k+1}(1)=0$ for odd $m=2k+1$ and $F_{2k}(1)=2C_{2k}(1)$ for even $m=2k$:

$$\begin{aligned} \begin{aligned} F_{2k+1}(1)&= -\sum _{j=0}^{2k+1}\, (-1)^j \widetilde{B}_{2j}\widetilde{B}_{4k+2-2j} = 0 , \\ F_{2k}(1)&= -\sum _{j=0}^{2k} \,(-1)^j \widetilde{B}_{2j}\widetilde{B}_{4k-2j} = 2 C_{2k}(1)= 2 \,\frac{\zeta (4k-1)+ E_{2k}(1)}{(2\pi )^{4k-1}} \hspace{0.55542pt}, \\ E_{2k}(1)&= \sum _{n=1}^\infty \, \frac{2}{n^{4k-1}\hspace{0.55542pt}(e^{2\pi n}-1)}\hspace{0.55542pt}. \end{aligned} \end{aligned}$$

(7)

Thus, with $F^A_{2k}=2^{4k-1}F_{2k}(1)$ we obtain for $\zeta (4k-1)$,

$$\begin{aligned} \begin{aligned} 2\hspace{0.55542pt}\zeta (4k-1)&= \pi ^{4k-1} F^A_{2k} - 2 E_{2k}(1) , \\ F^A_{2k}&={}-2^{4k-1} \sum _{j=0}^{2k}\, (- 1)^j \widetilde{B}_{2j}\widetilde{B}_{4k-2j} , \end{aligned} \end{aligned}$$

(8)

where $E_{2k}(1)$ converges very rapidly with n, namely $\propto e^{-2\pi n}$. The values of $F^A_m(1)$ are shown in Table 1.

(b) For $\alpha =2$ we get with (5):

$$\begin{aligned} F_m(2)&= -\sum _{j=0}^m \,(-1)^j 2^{2j-m} \widetilde{B}_{2j}\widetilde{B}_{2m-2j} = 2^{m-1} C_m\biggl (\frac{1}{2}\biggr ) - (-2)^{1-m} C_m(2) \\&= 2^{m-1}\hspace{1.111pt}\frac{\zeta (2m-1)+ E_m(1/2)}{(2\pi )^{2m-1}} - (- 2)^{1-m} \hspace{1.111pt}\frac{\zeta (2m-1)+ E_m(2)}{(2\pi )^{2m-1}}\hspace{0.55542pt}, \\ E_m\biggl (\frac{1}{2}\biggr )&= \sum _{n=1}^\infty \, \frac{2}{n^{2m-1}\, (e^{\pi n}-1)} , \quad E_m(2) = \sum _{n=1}^\infty \,\frac{2}{n^{2m-1}\hspace{0.55542pt}(e^{4\pi n}-1)}\hspace{0.55542pt}. \end{aligned}$$

Multiplying with $2^m$, and writing $F^B_m = 2^{3m-1} F_m(2)$, $\zeta (2m-1)$ is calculated by

$$\begin{aligned} \begin{aligned}&\bigl (u^B_m + w^B_m\bigr )\hspace{0.55542pt}\zeta (2m-1) = \pi ^{2m-1} F^B_m - u^B_m E_m\biggl (\frac{1}{2}\biggr ) - w^B_m E_m(2) ,\\&u^B_m = 2^{2m-1} , \quad w^B_m = 2(- 1)^m , \quad F^B_m = {}-2^{2m-1} \sum _{j=0}^m\, (- 4)^j \widetilde{B}_{2j}\widetilde{B}_{2m-2j} , \end{aligned} \end{aligned}$$

(9)

where the values of $u^B_m,w^B_m$, and $F^B_m$ are shown in Table 1.

(c) The periodicity of the $\coth$ function is given by $\coth \hspace{0.55542pt}(z+ki\pi )=\coth \hspace{0.55542pt}(z)$. Hence we conclude

$$\begin{aligned} C_m(\alpha +ki) = \sum _{n=1}^\infty \,\frac{\coth \hspace{0.55542pt}(\pi n (\alpha +ki))}{(2\pi n)^{2m-1}} = \sum _{n=1}^\infty \,\frac{\coth \hspace{0.55542pt}(\pi n \alpha )}{(2\pi n)^{2m-1}} = C_m(\alpha ) . \end{aligned}$$

(10)

Furthermore, with $\coth \hspace{0.55542pt}(z+i\pi /2)=\tanh \hspace{0.55542pt}(z)=2\coth \hspace{0.55542pt}(2z)-\coth \hspace{0.55542pt}(z)$ we obtain

$$\begin{aligned} \begin{aligned} C_m&\biggl (\alpha +\frac{i}{2}\biggr ) = \sum _{n=1}^\infty \frac{\coth \hspace{0.55542pt}(\pi n(\alpha +i/2))}{{(2\pi n)^{2m-1}}}\\&= \sum _{n=1}^\infty \frac{\coth \hspace{0.55542pt}(\pi n\alpha +i\pi /2)}{(2\pi n)^{2m-1}} + \sum _{n=1}^\infty \frac{\coth \hspace{0.55542pt}(2\pi n\alpha )-\coth \hspace{0.55542pt}(2\pi n\alpha +i\pi /2)}{(4\pi n)^{2m-1}} \\&= \sum _{n=1}^\infty \frac{ -\coth \hspace{0.55542pt}(\pi n\alpha ) + (2+2^{2-2m})\coth \hspace{0.55542pt}(2\pi n\alpha ) - 2^{2-2m}\coth \hspace{0.55542pt}(4\pi n\alpha )}{{(2\pi n)^{2m-1}}} \\&= {}- C_m(\alpha ) + (2+4^{1-m})\hspace{0.55542pt}C_m(2\alpha ) - 4^{1-m} C_m(4\alpha ) . \end{aligned} \end{aligned}$$

(11)

Now, for $\alpha =1+i$ we have $\alpha ^{-1}=(1-i)/2$. With (10) and (11) we get

$$\begin{aligned} C_m(1+i)&= C_m(1) , \\ C_m\biggl (\frac{1-i}{2}\biggr )&= {}- C_m\biggl (\frac{1}{2}\biggr ) + (2+4^{1-m})\hspace{0.55542pt}C_m(1) - 4^{1-m} C_m(2) , \end{aligned}$$

where $C_m(1+i)$ and $C_m((1-i)/2)$ are real numbers. Then, (5) gives

$$\begin{aligned} F_m&(1+i) =-\sum _{j=0}^m \,(-1)^j (1+i)^{2j-m} \widetilde{B}_{2j}\widetilde{B}_{2m-2j} \\&= (1+i)^{m-1} C_m\biggl (\frac{1-i}{2}\biggr ) - [- (1+i)]^{1-m}\hspace{0.55542pt}C_m(1+i) \\&= (1+i)^{m-1} \biggl [{-}\hspace{1.66656pt}C_m\biggl (\frac{1}{2}\biggr ) + \bigl (2+4^{1-m}-(- 2i)^{1-m}\bigr )\hspace{1.111pt}C_m(1) - 4^{1-m} C_m(2) \biggr ] . \end{aligned}$$

Defining $F^H_m=2^m (1-i)^{m-1} F_m(1+i)$, we have

$$\begin{aligned} F^H_m&= {}-\frac{1}{2}\, (1+i)(- 2i)^m \sum _{j=0}^m\, (- 2i)^j \widetilde{B}_{2j}\widetilde{B}_{2m-2j} \\&= {}-2^{2m-1} C_m\biggl (\frac{1}{2}\biggr ) + \bigl (2^{2m}+2-2(2i)^{m-1}\bigr )\hspace{1.111pt}C_m(1) - 2 C_m(2) . \end{aligned}$$

Now, looking at the real part, with $F^C_m= 2^{2\,m-1} \textrm{Re}\, F^H_m$ we obtain

$$\begin{aligned} \bigl (u^C_m&+ v^C_m + w^C_m\bigr ) \hspace{1.111pt}\zeta (2m-1) \\&= \pi ^{2m-1} F^C_m- u^C_m E_{m}\biggl (\frac{1}{2}\biggr ) - v^C_m E_{m}(1) - w^C_m E_{m}(2) , \nonumber \\ u^C_m&= {}-2^{2m-1} , \quad v^C_m = 2^{2m} + 2 - \textrm{Im}\hspace{0.55542pt}[(2i)^m], \quad w^C_m = {}-2 ,\nonumber \\ F^C_{2k}&= {}-\frac{1}{4}\, (-64)^{k} \biggl [ \sum _{j=0}^k \,(-4)^j \widetilde{B}_{4j}\widetilde{B}_{4k-4j} +2 \sum _{j=0}^{k-1}\, (-4)^j \widetilde{B}_{4j+2}\widetilde{B}_{4k-4j-2} \biggr ] ,\nonumber \\ F^C_{2k+1}&= {}-2 (-64)^{k} \biggl [ \sum _{j=0}^k (-4)^j \widetilde{B}_{4j}\widetilde{B}_{4k+2-4j} -2 \sum _{j=0}^{k} (-4)^j \widetilde{B}_{4j+2}\widetilde{B}_{4k-4j} \biggr ] .\nonumber \end{aligned}$$

(12)

The values of the coefficients $u^C_m,v^C_m,w^C_m$, and $F^C_m$ are presented in Table 1. Calculating the imaginary part, $\textrm{Im}\, F^H_m$, and comparing with (7), we find nontrivial relations of modified Bernoulli numbers:

$$\begin{aligned} {}-2\hspace{0.55542pt}(-4)^{-k} \hspace{0.55542pt}\textrm{Im}\, F^H_{2k}&= \sum _{j=0}^k\, (-4)^j \widetilde{B}_{4j}\widetilde{B}_{4k-4j} - 2\sum _{j=0}^{k-1} \,(-4)^j \widetilde{B}_{4j+2}\widetilde{B}_{4k-4j-2}\\&= 2 C_{2k}(1) = F_{2k}(1) = -\sum _{j=0}^{2k} \,(-1)^j \widetilde{B}_{2j}\widetilde{B}_{4k-2j} , \\ (-4)^{-k}\hspace{0.55542pt}\textrm{Im}\, F^H_{2k+1}&= \sum _{j=0}^k\, (-4)^j \widetilde{B}_{4j}\widetilde{B}_{4k+2-4j} + 2\sum _{j=0}^{k}\, (-4)^j \widetilde{B}_{4j+2}\widetilde{B}_{4k-4j} = 0 , \end{aligned}$$

see also Vepstas [13].

(d) In (12) the sums $E_m(1/2) \,{\propto }\, \exp \hspace{0.55542pt}(-n\pi )$ do not converge as fast as the sums $E_m(1) \,{\propto }\, \exp \hspace{0.55542pt}(-2n\pi )$ and $E_m(2) \,{\propto }\, \exp \hspace{0.55542pt}(-4n\pi )$. The sums $E_m(1/2)$, which occur in $F^B_m$, (9), and $F^C_m$, (12), can be avoided by taking the linear combination

$$\begin{aligned} F^D_m = F^B_m + F^C_m. \end{aligned}$$

Since $u^B_m+u^C_m=2^{2m-1}-2^{2m-1}=0$, we get

$$\begin{aligned} \begin{aligned}&\bigl (v^D_m + w^D_m\bigr )\hspace{1.111pt}\zeta (2m-1) = \pi ^{2m-1} F^D_m - v^D_m E_{m}(1) - w^D_m E_{m}(2) , \\&v^D_{2k} = 16^k + 2 , \quad w^D_{2k} = 0 , \quad F^D_{2k} = F^B_{2k} + F^C_{2k} , \\&v^D_{2k+1} = 4^{2k+1} + 2 - 2(-4)^k , \quad w^D_{2k+1} = -4 , \quad F^D_{2k+1} = F^B_{2k+1} + F^C_{2k+1} . \end{aligned} \end{aligned}$$

(13)

For the values of $v^D_m,w^D_m$, and $F^D_m$, see Table 1. For even $m=2k$ we get $w^D_{2k}=0$. Thus, comparing (13) with (8), $F^D_{2k} = F^B_{2k}+ F^C_{2k} = (2^{4k-1}+1) F^A_{2k}$ gives another relation of modified Bernoulli numbers $\widetilde{B}_{2j}$:

$$\begin{aligned} (-4)^{k}&\,\biggl [ \sum _{j=0}^k \,(-4)^j \widetilde{B}_{4j}\widetilde{B}_{4k-4j} +2 \sum _{j=0}^{k-1} \,(-4)^j \widetilde{B}_{4j+2}\widetilde{B}_{4k-4j-2} \biggr ] \\&= (16^k+2) \sum _{j=0}^{2k} \,(-1)^j \widetilde{B}_{2j}\widetilde{B}_{4k-2j} - 2 \sum _{j=0}^{2k} \,(-4)^j \widetilde{B}_{2j}\widetilde{B}_{4k-2j} . \end{aligned}$$

Calculating $\zeta (2m-1)$, the series of (8) for even m and of (13) for odd m converge with $\propto \exp \hspace{0.55542pt}(-2n\pi )$ faster than the series of (9) and (12) with $\propto \exp \hspace{0.55542pt}(-n\pi )$.

(e) If $F^E_m$ is a linear combination of $F^A_m,F^B_m$, and $F^C_m$, see (8), (9), (12), such that $u^E_m+v^E_m+w^E_m=0$, then the contribution of $\zeta (2m-1)$ to $F^E_m$ vanishes, and $(2\pi )^{2m-1} F^E_m$ can be calculated solely by the sums $E_m(1/2),E_m(1)$, and $E_m(2)$, compare also Vepstas [13]. Thereby, $\pi ^{2m-1}$ can be expressed by fast converging series. For m up to 10, these sums are presented by Plouffe [9]. In general, they can be achieved by the following linear combinations:

$$\begin{aligned}\begin{aligned} F^E_{2k}&= F^B_{2k} - \bigl (1 +4^{2k-1}\bigr )\hspace{0.55542pt}F^A_{2k}{} & {} \text{ for } \text{ even } m=2k , \\ F^E_{2k+1}&= (-4)^k F^B_{2k+1} - \bigl [(-4)^k +1\bigr ]\hspace{0.55542pt}F^C_{2k+1}{} & {} \text{ for } \text{ odd } m=2k+1 , \end{aligned} \end{aligned}$$

where we get

(14)

The values of $u^E_m,v^E_m$, $w^E_m$, and $F^E_m$ are listed in Table 1. Because the sums $E_m(1/2)$, $E_m(1)$, and $E_m(2)$ themselves depend on $e^{\pi }$, a calculation of $\pi$ by (14) must be carried out numerically.

Table 1 The coefficients of (a) (8), (b) (9), (c) (12), (d) (13), and (e) (14), for $2 \leqslant m \leqslant 8$

3 The limit $\alpha \rightarrow 0$

The functions $F_m(\alpha )$ of (1) do not only yield series for calculating $\zeta (2m-1)$. Additionally, the limits $\lim _{\,\alpha \rightarrow 0} \alpha ^m F_m(\alpha )$ in (1) derive Euler’s classical formula for $\zeta (2m)$.

Theorem 3.1

$$\begin{aligned} \frac{(-1)^m \hspace{0.55542pt}(2\pi )^{2m}}{2} \lim _{\alpha \rightarrow 0} \alpha ^m F_m(\alpha ) = \zeta (2m) = \sum _{n=1}^\infty \frac{1}{n^{2m}} = \frac{(-1)^{m-1}\hspace{0.55542pt}(2\pi )^{2m}}{2\hspace{0.55542pt}(2m)!} \,B_{2m}. \end{aligned}$$

Notably, the analytical expressions of even zeta values and the fast-converging series of odd zeta values arise from the same set of functions $F_m(\alpha )$.

Proof

(a) $m\geqslant 2$: Using the identities

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \pi n\alpha \coth \hspace{0.55542pt}(\pi n\alpha )=1, \quad \lim _{\alpha \rightarrow 0} \alpha ^{2m-1}\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})=0, \end{aligned}$$

and $\widetilde{B}_0=1$, see (2), the limit $\lim _{\,\alpha \rightarrow 0} \alpha ^m F_m(\alpha )$ for both sides of (1) gives

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \alpha ^m F_m(\alpha )&= \lim _{\alpha \rightarrow 0} \sum _{j=0}^m \,(-1)^{j+1} \alpha ^{2j} \widetilde{B}_{2j}\widetilde{B}_{2m-2j} = {}-\widetilde{B}_{2m} = {}-\frac{B_{2m}}{(2m)!} \\&= \lim _{\alpha \rightarrow 0}\, \biggl [\, \sum _{n=1}^\infty \, \frac{\alpha ^{2m-1}\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})}{(2\pi n)^{2m-1}} +(-1)^m \sum _{n=1}^\infty \,\frac{\alpha \coth \hspace{0.55542pt}(\pi n\alpha )}{(2\pi n)^{2m-1}} \biggr ]\\&= \sum _{n=1}^\infty \, \frac{2(-1)^m}{(2\pi n)^{2m}}\hspace{0.55542pt}, \end{aligned}$$

(b) $m=1$: Analogously to (1), for $m=1$ the function $F_1(\alpha )$ is calculated as

$$\begin{aligned} F_1(\alpha ) = C_1(\alpha ^{-1}) - C_1(\alpha ) = \sum _{n=1}^\infty \frac{\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})-\coth \hspace{0.55542pt}(\pi n\alpha )}{2\pi n} = \frac{\alpha -\alpha ^{-1}}{12} - \frac{\ln \hspace{0.55542pt}(\alpha )}{2\pi }\hspace{0.55542pt}, \end{aligned}$$

see e.g. [2, 8]. With $\lim _{\,\alpha \rightarrow 0} \alpha \ln \hspace{0.55542pt}(\alpha )=0$ and $\widetilde{B}_2=1/12$, see (2), here the limit $\lim _{\,\alpha \rightarrow 0} \alpha F_1(\alpha )$ results

$$\begin{aligned} \lim _{\alpha \rightarrow 0} \alpha F_1(\alpha )&\displaystyle = \lim _{\alpha \rightarrow 0}\hspace{1.111pt}\alpha \biggl [ \frac{\alpha -\alpha ^{-1}}{12} - \frac{\ln \hspace{0.55542pt}(\alpha )}{2\pi } \biggr ] = {}-\frac{1}{12} = {}-\widetilde{B}_{2} = {}-\frac{B_{2}}{2!}\nonumber \\&= \lim _{\alpha \rightarrow 0}\, \sum _{n=1}^\infty \,\alpha \, \frac{\coth \hspace{0.55542pt}(\pi n\alpha ^{-1})-\coth \hspace{0.55542pt}(\pi n\alpha )}{2\pi n} = \sum _{n=1}^\infty \, \frac{-2}{(2\pi n)^2}\hspace{0.55542pt}. \end{aligned}$$

$\square$

4 Triangle identity

Whereas $F_m(\alpha )=\alpha ^{m-1}C_m(\alpha ^{-1})-(-\alpha ^{-1})^{1-m}C_m(\alpha )$ can be calculated directly by the coefficients $\widetilde{B}_{2j}$, see (5), no closed-form solutions are known for the sums $C_m(\alpha )$, apart from those $\alpha _k$, where $\alpha _k=\alpha _k^{-1}+ki$ and therefore $C_m(\alpha _k)=C_m(\alpha _k^{-1})$ due to symmetry. Moreover, the function values $F_m(\alpha ),F_m(\alpha +i)$, and $F_m(\alpha ^{-1}-i)$ form a triangle identity.

Theorem 4.1

$$\begin{aligned} F_m(\alpha ) = (i\alpha ^{-1}+1)^{m-1} F_m(\alpha +i) - (i\alpha -1)^{m-1} F_m(\alpha ^{-1}-i). \end{aligned}$$

Proof

With the symmetry of (10), $C_m(\alpha )= C_m(\alpha \pm i)$, we have

$$\begin{aligned} C_m(\alpha ) = C_m(\alpha +i), \;\; C_m\biggl (\frac{1}{\alpha }\biggr ) = C_m\biggl (\frac{1-i\alpha }{\alpha }\biggr ), \;\; C_m\biggl (\frac{1}{\alpha +i}\biggr ) = C_m\biggl (\frac{\alpha }{1-i\alpha }\biggr ). \end{aligned}$$

Then, with (5) we get

which proves the theorem. $\square$

Analogously, the generating function $f_\alpha (x)=-\sum _{m=2}^\infty F_m(\alpha )x^{2m}$, (4), shows a similar symmetry, satisfying the relation

$$\begin{aligned} 0&= f_\alpha (x) - (i\alpha ^{-1}+1)^{-1} f_{\alpha +i}\bigl (\sqrt{i\alpha ^{-1}+1}\, x\bigr ) + (i\alpha -1)^{-1} f_{\alpha ^{-1}-i}\bigl (\sqrt{i\alpha -1}\, x\bigr ) \\&= -\sum _{m=2}^\infty x^{2m} \bigl [ F_m(\alpha ) - (i\alpha ^{-1}+1)^{m-1} F_m(\alpha +i) + (i\alpha -1)^{m-1} F_m(\alpha ^{-1}-i) \bigr ] . \end{aligned}$$