The following formula was stated by Ramanujan:

$$\sum\limits_{n=1}^{\infty}\frac{\coth n\pi}{n^7}=\frac{19\pi^7}{56700}$$

Does anybody know the method of proof of this formula? I know that typically Ramanujan used extensively methods of divergent series, but I cannot see how to attempt a proof of this result. It looks somehow like a relatively simple result, but I can't see what methods might be used to obtain it.


Since $(7)$ from this answer is valid for any $z\in\mathbb{C}$, we have $$ \begin{align} \pi\coth(\pi n) &=\sum_{k\in\mathbb{Z}}\frac1{n+ik}\\ &=\frac1n+2n\sum_{k=1}^\infty\frac1{n^2+k^2}\tag{1} \end{align} $$ Therefore, $$ \begin{align} \sum_{n=1}^\infty\frac{\pi\coth(\pi n)}{n^7} &=\sum_{n=1}^\infty\frac1{n^8}+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^6(n^2+k^2)}\\ &=\zeta(8)+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\left(\frac1{n^6}-\frac1{n^4(n^2+k^2)}\right)\\ &=\zeta(8)+2\zeta(2)\zeta(6)-2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\frac1{n^4(n^2+k^2)}\tag{2}\\ &=\zeta(8)+2\zeta(2)\zeta(6)-2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^4}\left(\frac1{n^4}-\frac1{n^2(n^2+k^2)}\right)\\ &=\zeta(8)+2\zeta(2)\zeta(6)-2\zeta(4)\zeta(4)+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^4n^2(n^2+k^2)}\tag{3}\\[6pt] &=\zeta(8)+2\zeta(2)\zeta(6)-\zeta(4)\zeta(4)\tag{4}\\[12pt] &=\frac{19\pi^8}{56700}\tag{5} \end{align} $$ where $(4)$ is the average of $(2)$ and $(3)$. Also. we've used the values of $\zeta(2k)$ computed in this answer. Thus, $$ \sum_{n=1}^\infty\frac{\coth(\pi n)}{n^7}=\frac{19\pi^7}{56700}\tag{6} $$


Suppose we seek to show that $$\sum_{n\ge 1} \frac{\coth(n \pi)}{n^7} = \frac{19\pi^7}{56700}.$$

Using $$\coth(x) = \frac{e^x+e^{-x}}{e^x-e^{-x}} = 1 + 2\frac{e^{-x}}{e^x-e^{-x}}$$ this is the same as $$2\sum_{n\ge 1} \frac{1}{n^7} \frac{e^{-n\pi}}{e^{n\pi}-e^{-n\pi}} = -\zeta(7) + \frac{19\pi^7}{56700}.$$

The sum term may be evaluated using harmonic summation techniques. Since this method has not been presented I will detail this calculation here.

Put $$S(x) = \sum_{n\ge 0} \frac{1}{n^7} \frac{e^{-nx}}{e^{nx}-e^{-nx}}.$$

We will evaluate $S(\pi)$ using a functional equation for $S(x)$ that is obtained by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^7}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{e^{-x}}{e^x-e^{-x}}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{e^{-x}}{e^x-e^{-x}} x^{s-1} dx = \int_0^\infty \frac{e^{-2x}}{1-e^{-2x}} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 0} e^{-2x} e^{- 2 q x} x^{s-1} dx = \sum_{q\ge 0} \int_0^\infty e^{-2(q+1)x} x^{s-1} dx \\= \Gamma(s) \sum_{q\ge 0} \frac{1}{2^s (q+1)^s} = \frac{1}{2^s} \Gamma(s) \zeta(s)$$ with fundamental strip $\langle 1, \infty\rangle.$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = 2^{-s} \Gamma(s) \zeta(s) \zeta(s+7) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^7} \frac{1}{k^s}$$ for $\Re(s) > -6.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

Fortunately the trivial zeros of the two zeta function terms cancel the poles of the gamma function term. Shifting to $\Re(s) = -7 -1/2$ we get $$S(x) = \frac{\pi^8}{18900}\frac{1}{x} - \frac{1}{2} \zeta(7) + \frac{\pi^6 x}{5670} - \frac{\pi^4 x^3}{8100} + \frac{\pi^2 x^5}{5670} + \frac{4}{45} \zeta'(-6) x^6 + \frac{1}{18900} x^7 \\+ \frac{1}{2\pi i} \int_{-15/2-i\infty}^{-15/2+i\infty} Q(s)/x^s ds.$$

We will turn this into the promised functional equation.

Substitute $s = -6 - t$ in the remainder integral to get $$- \frac{1}{2\pi i} \int_{3/2+i\infty}^{3/2-i\infty} \frac{1}{2^{-6-t}} \Gamma(-6-t) \zeta(-6-t) \zeta(1-t) x^{t+6} dt$$ which is $$\frac{x^6}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} 2^{6+t} \Gamma(-6-t) \zeta(-6-t) \zeta(1-t) x^t dt$$

In view of the desired functional equation we now use the functional equation of the Riemann zeta function on $Q(s)$ to prove that the integrand of the last integral is in fact $-Q(t)/\pi^{6+2t}.$

Start with the functional equation $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ and substitute this into $Q(s)$ to obtain $$Q(s) = 2^{-s} \frac{\zeta(1-s) 2^s \pi^s}{2\cos\left(\frac{\pi s}{2}\right)} \zeta(s+7) = \frac{1}{2} \pi^s \frac{\zeta(s+7)}{\cos\left(\frac{\pi s}{2}\right)} \zeta(1-s).$$ Apply the functional equation again (this time to $\zeta(s+7)$) to get $$Q(s) = \frac{1}{2} \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)} \frac{2}{2^{-6-s} \pi^{-6-s}} \cos\left(\frac{\pi (-6-s)}{2}\right) \Gamma(-6-s) \zeta(-6-s) \zeta(1-s)$$ Observe that $$\frac{\cos\left(-3\pi-\frac{\pi s}{2}\right)} {\cos\left(\frac{\pi s}{2}\right)} = - \frac{\cos\left(-\frac{\pi s}{2}\right)} {\cos\left(\frac{\pi s}{2}\right)} = -1$$ so we finally get $$Q(s) = - 2^{6+s} \pi^{6+2s} \Gamma(-6-s) \zeta(-6-s) \zeta(1-s),$$ thus proving the claim.

Return to the remainder integral and re-write it as follows: $$\frac{(x/\pi)^6}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} 2^{6+t} \pi^{6+2t} \Gamma(-6-t) \zeta(-6-t) \zeta(1-t) (x/\pi^2)^t dt.$$ so that the fact of it being a multiple of the defining integral of $S(x)$ becomes readily apparent.

Using the fact that $4/45 \times \zeta'(-6) = -1/2\times\zeta(7)/\pi^6$ we have established the functional equation $$S(x) = \frac{\pi^8}{18900}\frac{1}{x} - \frac{1}{2} \zeta(7) + \frac{\pi^6 x}{5670} - \frac{\pi^4 x^3}{8100} + \frac{\pi^2 x^5}{5670} - \zeta(7) \frac{1}{2\pi^6} x^6 + \frac{x^7}{18900} \\ - \frac{x^6}{\pi^6} S(\pi^2/x).$$

Now the value $x=\pi$ is obviously special here and we get $$S(\pi) = \pi^7 \left(\frac{1}{18900} + \frac{1}{5670} - \frac{1}{8100} + \frac{1}{5670} + \frac{1}{18900}\right) -\zeta(7)- S(\pi)$$ which gives $$2 S(\pi) = \pi^7 \frac{19}{56700} -\zeta(7)$$ as was to be shown.

The inspiration for this calculation is from the paper "Mellin Transform and its Applications" by Szpankowski.

  • 1
    $\begingroup$ Great! I'm ignorant of these techniques though. $\endgroup$ – Hobbyist Oct 28 '14 at 8:50
  • $\begingroup$ Nicely done. Is there a deeper reason why S(pi) is as special as it is or does that just happen to be that way? $\endgroup$ – WorldSEnder Jul 4 '18 at 6:41

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There is a straightforward way to evaluate the $\texttt{initial (k,n)-sum}$ which is the starting point of @robjohn fine answer:

\begin{align} \sum_{n = 1}^{\infty}{\pi\coth\pars{\pi n} \over n^{7}} & = \sum_{n = 1}^{\infty}{1 \over n^{8}} + 2\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}{1 \over n^{6}\pars{n^{2} + k^{2}}} \\[5mm] & = \zeta\pars{8} + \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}\bracks{% {1 \over n^{6}\pars{n^{2} + k^{2}}} + {1 \over k^{6}\pars{k^{2} + n^{2}}}} \\[5mm] & = \zeta\pars{8} + \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty} {k^{6} + n^{6} \over k^{6}n^{6}\pars{n^{2} + k^{2}}} = \zeta\pars{8} + \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty} {k^{4} - k^{2}n^{2} + n^{4} \over k^{6}n^{6}} \\[5mm] & = \zeta\pars{8} + \sum_{n = 1}^{\infty}{1 \over n^{6}}\sum_{k = 1}^{\infty}{1 \over k^{2}} - \sum_{n = 1}^{\infty}{1 \over n^{4}}\sum_{k = 1}^{\infty}{1 \over k^{4}} + \sum_{n = 1}^{\infty}{1 \over n^{2}}\sum_{k = 1}^{\infty}{1 \over k^{6}} \\[5mm] & = \zeta\pars{8} + 2\zeta\pars{6}\zeta\pars{2} - \zeta^{2}\pars{4} \end{align}


In the following a general expression for $$ \sum_{n=1}^\infty\frac{\coth(n\pi)}{n^{4K-1}} $$ with $K\in\mathbb Z_+$ will be derived.

Consider the following function: $$ f_k(z)=\frac{\cot z\coth z}{z^{k}}.\tag1 $$ The function has the pole of order $k+2$ at $z=0$ and simple poles at the points $z=n\pi$ and $z=i n\pi$, with $n\in\mathbb Z$, $n\ne0$.

Let us integrate the function along a square contour connecting the following points of the complex plane: $$\gamma_\nu:\; (-\nu,-\nu)\to(\nu,-\nu)\to(\nu,\nu)\to(-\nu,\nu)\to $$ with $\nu\in\mathbb R,\nu\not\in\mathbb Z,\nu>0$.

By Cauchy's theorem the exact choice of $\nu\in(n\pi,n\pi+\pi)$ does not matter for the value of the integral. To simplify its estimate for large $n$ let $\nu=\left(n+\frac14\right)\pi$. In this case: $$ |\cot(x\pm i\nu)\coth(x\pm i\nu)|^2=\frac{1+\cot^2x\coth^2\nu}{\cot^2x+\coth^2\nu}\cdot \underbrace{\frac{1+\coth^2x\cot^2\nu}{\coth^2x+\cot^2\nu}}_{=1} =\frac{\tan^2x+\coth^2\nu}{1+\tan^2x\coth^2\nu}\le\coth^2\nu. $$ Due to the symmetry the same estimate is valid for $|\cot(i x\pm \nu)\coth(i x\pm \nu)|^2$. Therefore: $$ \lim_{n\to\infty}\oint_{\gamma_{\nu}}f_k(z)dz=0. $$ Thus by the residue theorem we have: $$ \operatorname{Res}(f_k,0)+\sum_{n=1}^\infty\left[\operatorname{Res}(f_k,n\pi)+\operatorname{Res}(f_k,-n\pi)+\operatorname{Res}(f_k,in\pi)+\operatorname{Res}(f_k,-in\pi)\right]=0\tag2 $$

The pole at $z=0$ can be calculated as follows. Recall that: $$ \cot z=\frac1z\sum_{n=0}^\infty\frac{(-1)^nB_{2n}}{(2n)!}(2z)^{2n};\quad \coth z=\frac1z\sum_{n=0}^\infty\frac{B_{2n}}{(2n)!}(2z)^{2n}, $$ where $B_n$ are Bernoulli numbers.

From this one obtains: $$ \cot z\coth z=\frac1{z^2}\sum_{n=0}^\infty\frac{A_{n}}{n!}(2z)^{n},\text{ with } A_{2n+1}=0,\;A_{2n}=\sum_{k=0}^n(-1)^k\binom{2n}{2k}B_{2k}B_{2n-2k}.\tag3 $$ It follows $$ \operatorname{Res}(f_k,0)=\frac{2^{k+1}A_{k+1}}{(k+1)!}\equiv A^*_{k+1}.\tag4 $$

Next we compute the residues at the simple poles: $$ \operatorname{Res}(f_k,n\pi)=\lim_{z\to n\pi}(z-n\pi)\frac{\cot z\coth z}{z^{k}}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta)\coth(\zeta+n\pi)}{(\zeta+n\pi)^{k}}=\frac{\coth(n\pi)}{(n\pi)^{k}},\tag{5a} $$ and $$ \operatorname{Res}(f_k,i n\pi)=\lim_{z\to in\pi}(z-in\pi)\frac{\cot z\coth z}{z^{k}}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta+in\pi)\coth(\zeta)}{(\zeta+i n\pi)^{k}}=\frac{\coth(n\pi)}{i(in\pi)^{k}},\tag{5b} $$ where we used $$\begin{align} &\lim_{x\to0}x\cot x=\lim_{x\to0}x\coth x=1,\\ &\cot(x+n\pi)=\cot(x),\\ &\coth(x+in\pi)=\coth(x),\\ &\cot(i x)=-i\coth(x). \end{align} $$

From $(5)$ one obtains: $$ \operatorname{Res}(f_k,n\pi)+\operatorname{Res}(f_k,-n\pi)+\operatorname{Res}(f_k,in\pi)+\operatorname{Res}(f_k,-in\pi) =\begin{cases} 4\dfrac{\coth(n\pi)}{(n\pi)^k},&k=-1\text{ mod }4,\\ 0,&k\ne -1\text{ mod }4.\\ \end{cases}\tag6 $$

Combining $(2)$, $(4)$ and $(6)$ one draws two conclusions: $$A_k=0\text{ for } k\ne 0\text{ mod }4.\tag7$$ $$4\sum_{n=1}^\infty\dfrac{\coth(n\pi)}{(n\pi)^{4K-1}}=-A^*_{4K}.\tag8$$

Observe that (7) is non-trivial only for $k=2\text{ mod }4$. No closed-form expression is known for $A_{4K}$ but its calculation using $(3)$ represents no difficulty. The first four values of $A^*_{4K}$ are: $$ A^*_0=1,\quad A^*_4=-\frac7{45},\quad A^*_8=-\frac{19}{14175},\quad A^*_{12}=-\frac{2906}{212837625}. $$

Particularly, for $K=2$ one obtains: $$\sum_{n=1}^\infty\dfrac{\coth(n\pi)}{n^{7}}=-\dfrac{A^*_{8}\pi^7}4 =\dfrac{19\pi^7}{56700}.$$

Using for $n\ne0$ the relation: $$ B_{2n}=\frac{(-1)^{n+1}(2n)!}{(2\pi)^{2n}}2\zeta(2n), $$ where $\zeta(z)$ is the Riemann zeta-function, the equation $(8)$ can be rewritten as: $$ \pi\sum_{n=1}^\infty\dfrac{\coth(n\pi)}{n^{4K-1}}= \zeta(4K)-\sum_{k=1}^{2K-1}(-1)^k\zeta(2k)\zeta(4K-2k), $$ in agreement with previous results for $K=2$.


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