# A Ramanujan sum involving $\sinh$

Today, in a personal communication, I was asked to prove the classical result $$\boxed{ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)} = \frac{\pi^3}{360}}\tag{CR}$$ which I believe is due to Ramanujan. My proof can be found here and it is based on the closed form for $\sum_{n\geq 1}\frac{1}{m^2+n^2}$ and double counting. I would use this question for collecting alternative proofs. I am already aware that standard techniques for tackling series with a similar structure are

1. The Poisson summation formula;
2. The residue theorem coupled with the Laplace transform of $\text{Li}_k$ (I used this approach due to Simon Plouffe here, for instance);
3. Eisenstein series related to Gaussian integers;
4. Identities involving Dirichlet's series, since $$r_2(N)=\left|\{(a,b)\in\mathbb{Z}^2:a^2+b^2=N\}\right|=4 \sum_{d\mid N}\chi_4(d)$$ (this can be regarded as an analytic-combinatorial equivalent of the statement "$\mathbb{Z}[i]$ is a UFD").

I also know that a cornerstone is given by Zucker's The Summation of Series of Hyperbolic Functions, 1979. Let us see if we can devise a very short proof of $(\text{CR})$ through these ingredients or other ones.

• A folklore note: a straightforward consequence of $(\text{CR})$ is $\pi^3\approx 31$. – Jack D'Aurizio Jan 9 '18 at 16:40
• Thanks for that website MathB for sharing math content (like pastebin) . – Paramanand Singh Jan 10 '18 at 7:08

Suppose we seek to show that $$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{1}{\sinh(\pi n)} = \frac{\pi^3}{360}.$$

Using $$\frac{1}{\sinh(x)} = \frac{2}{e^x-e^{-x}} = 2\frac{e^{-x}}{1-e^{-2x}}$$ this is the same as

$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{e^{-n\pi}}{1-e^{-2n\pi}} = \frac{\pi^3}{720}.$$

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

Let $$p$$ be a positive integer and introduce

$$S(x;p) = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{2p+1}} \frac{e^{-nx}}{1-e^{-2nx}}.$$

We will evaluate $$S(\pi;p)$$ using a functional equation for $$S(x;p)$$ 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 $$\lambda_k = \frac{(-1)^{k+1}}{k^{2p+1}}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{e^{-x}}{1-e^{-2x}}.$$

We need the Mellin transform $$g^*(s)$$ of $$g(x)$$ which is

$$\int_0^\infty \frac{e^{-x}}{1-e^{-2x}} x^{s-1} dx = \int_0^\infty \sum_{q\ge 0} e^{- (2 q+1) x} x^{s-1} dx = \sum_{q\ge 0} \int_0^\infty e^{-(2q+1)x} x^{s-1} dx \\= \Gamma(s) \sum_{q\ge 0} \frac{1}{(2q+1)^s} = \left(1-\frac{1}{2^s}\right) \Gamma(s) \zeta(s)$$ with fundamental strip $$\langle 1, \infty\rangle.$$

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

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

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. The first term cancels those at $$-2m$$ where $$m\ge 1$$ and the second one the odd ones from $$-2p-3$$ on, which leaves the poles at $$s=1$$, and $$-2q-1$$ with $$0\le q\le p.$$ It would appear there is a pole at $$s=-2p$$ but this is not the case since we have two simple poles among the five terms but also two zero values, making for cancelation. The pole at $$s=0$$ is canceled as well.

For the residue at $$s=1$$ we find

$$\frac{1}{2} \frac{2^{2p+1}-1}{2^{2p+1}} \times 1 \times \zeta(2p+2) \frac{1}{x} = \frac{2^{2p+1}-1}{2^{2p+2}} \frac{(-1)^{p} B_{2p+2} (2\pi)^{2p+2}}{2 (2p+2)!} \frac{1}{x} \\ = (2^{2p+1}-1) \frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}.$$

The negative odd values at $$s=-2q-1$$ yield

$$\left(1-\frac{1}{2^{-2q-1}}\right) \left(1-\frac{1}{2^{2p-2q-1}}\right) \frac{(-1)^{2q+1}}{(2q+1)!} \zeta(-2q-1) \zeta(2p-2q) x^{2q+1} \\ = (1-2^{2q+1}) \left(1-\frac{1}{2^{2p-2q-1}}\right) \frac{1}{(2q+1)!} \frac{B_{2q+2}}{2q+2} \frac{(-1)^{p-q+1} B_{2p-2q} (2\pi)^{2p-2q}}{2(2p-2q)!} x^{2q+1} \\ = \frac{1}{2} (1-2^{2q+1}) (2^{2p-2q-1}-1) \frac{(-1)^{p-q+1}}{(2q+1)!} \frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)} x^{2q+1}.$$

Shifting to $$\Re(s) = -2p -3/2$$ we get

$$S(x;p) = (2^{2p+1}-1) \frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x} \\ + \frac{1}{2} \sum_{q=0}^p (1-2^{2q+1}) (2^{2p-2q-1}-1) \frac{(-1)^{p-q+1}}{(2q+1)!} \frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)} x^{2q+1} \\ + \frac{1}{2\pi i} \int_{-2p-3/2-i\infty}^{-2p-3/2+i\infty} Q(s)/x^s ds.$$

We will turn this into the promised functional equation.

Substitute $$s = -2p - t$$ in the remainder integral to get

$$- \frac{1}{2\pi i} \int_{3/2+i\infty}^{3/2-i\infty} \left(1-2^{2p+t}\right) \left(1-2^t\right) \Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t) x^{t+2p} dt$$ which is

$$\frac{x^{2p}}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \left(1-2^{2p+t}\right) \left(1-2^t\right) \Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t) x^{t} dt.$$

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 $$(-1)^p Q(t)/\pi^{2p+2t}.$$

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

$$Q(s) = \left(1-\frac{1}{2^s}\right) \left(1-\frac{1}{2^{s+2p}}\right) \frac{\zeta(1-s) 2^s \pi^s}{2\cos\left(\frac{\pi s}{2}\right)} \zeta(s+2p+1) \\ = \frac{1}{2} (2^s-1) \left(1-\frac{1}{2^{s+2p}}\right) \pi^s \frac{\zeta(s+2p+1)}{\cos\left(\frac{\pi s}{2}\right)} \zeta(1-s).$$

Apply the functional equation again (this time to $$\zeta(s+2p+1)$$) to get

$$Q(s) = \frac{1}{2} \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)} (2^s-1) \left(1-\frac{1}{2^{s+2p}}\right) \frac{2}{2^{-2p-s} \pi^{-2p-s}} \cos\left(\frac{\pi (-2p-s)}{2}\right) \\ \times \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s) \\ = \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)} (2^s-1) (2^{2p+s}-1) \pi^{2p+s} (-1)^p \cos\left(\frac{-\pi s}{2}\right) \\ \times \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)$$

and we finally get $$Q(s) = (-1)^p \pi^{2p+2s} (1-2^s) (1-2^{2p+s}) \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)$$

thus proving the claim.

$$(-1)^p \frac{(x/\pi)^{2p}}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} (-1)^p \pi^{2p+2t} \left(1-2^{2p+t}\right) \left(1-2^t\right) \\ \times \Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t) (x/\pi^2)^{t} dt.$$

so that the fact of it being a multiple of the defining integral of $$S(\pi^2/x; p)$$ becomes readily apparent.

We have established the functional equation

$$\color{green} {S(x;p) = (2^{2p+1}-1) \frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x} \\ + \frac{1}{2} \sum_{q=0}^p (1-2^{2q+1}) (2^{2p-2q-1}-1) \frac{(-1)^{p-q+1}}{(2q+1)!} \frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)} x^{2q+1} \\ + (-1)^p \left(\frac{x}{\pi}\right)^{2p} S(\pi^2/x;p).}$$

Now the value $$x=\pi$$ is obviously special here (fixed point) and we get for $$p=2r+1$$ with $$r\ge 0$$ ($$p$$ even yields a Bernoulli number identity)

$$S(\pi; 2r+1) = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{4r+3}} \frac{e^{-n\pi}}{1-e^{-2n\pi}} = - \frac{\pi^{4r+3}}{4} (2^{4r+3}-1) \frac{B_{4r+4}}{(4r+4)!} \\ + \frac{\pi^{4r+3}}{4} \sum_{q=0}^{2r+1} (1-2^{2q+1}) (2^{4r+1-2q}-1) \frac{(-1)^{q}}{(2q+1)!} \frac{B_{2q+2} B_{4r+2-2q}}{(4r+2-2q)! (q+1)}.$$

We obtain a rational multiple of $$\pi^{4r+3}.$$ Scale by two to get for

$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{1}{\sinh(\pi n)}, \quad \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^7} \frac{1}{\sinh(\pi n)}, \\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{11}} \frac{1}{\sinh(\pi n)}, \quad \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{15}} \frac{1}{\sinh(\pi n)}, \quad \ldots$$

the values

$$\bbox[5px,border:2px solid #00A000]{ {\frac {{\pi }^{3}}{360}},\quad {\frac {13\,{\pi }^{7}}{453600}},\quad {\frac {4009\,{\pi }^{11}}{13621608000}},\quad {\frac {13739\,{\pi }^{15}}{4547140416000}},\quad \ldots}$$

These are dominated by the first term

$$\frac{1}{\sinh(\pi)}.$$

• Your favorite technique again. +1 the formula (functional equation) in green was given by Ramanujan without proof and is the key to obtaining such sums. Unfortunately such sums can't be handled by translating the series into elliptic integrals. – Paramanand Singh Jan 10 '18 at 5:16
• The distinguishing feature of this problem are the symmetries in the transform $Q(s)$ which were quite special to observe. – Marko Riedel Jan 10 '18 at 5:47
• Hello Marko, I have a question about the evaluation of the integral using the residue theorem. You state which we evaluate by shifting it to the left for an expansion about zero which I understand as you close the contour on the left at $-\infty$. But for that $$Q(s)x^{-s}$$ must vanish as we go to $-\infty$. But as far as I'm concerned $\Gamma(s)\zeta(s)^2 \, x^{-s}$ does not vanish for any $x>0$ as we go to $-\infty$. So I would guess this is at most an asymptotic series. However what justifies this? – Diger Sep 1 '18 at 21:31
• We are not shifting the left side to minus infinity for an asymptotic series, we just shift it to pick up all the residues and the result after shifting is seen to be a multiple of the original integral by a substitution, making for an exact formula. – Marko Riedel Sep 2 '18 at 11:55
• Marko's result can be transformed into a simpler form. Please see the following link. oeis.org/A330905 – TOM May 2 at 1:33

This can be derived very simply using contour integration in the complex plane. Consider the contour integral

$$\pi \oint_C \frac{dz}{z^3 \sinh{(\pi z)} \sin{(\pi z)}}$$

where $C$ is a square centered at the origin of side $2 N+1$. As $N \to \infty$, one may show that the contour integral approaches zero. (Consider the magnitude of the integrand over the sides of the square.)

This means that the sum of the residues at the poles $z=\pm n$ and $z=\pm i n$ vanishes. Note that for $n \ne 0$, the residues at $z=n$ and $z=i n$ are equal to $(-1)^n/(\pi n^3 \sinh{(\pi n)})$. Also, the summand is even in $n$. Thus we have $$4 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3 \sinh{(\pi n)}} + \operatorname*{Res}_{z=0} \frac{\pi}{z^3 \sinh{(\pi z)} \sin{(\pi z)}} = 0$$

Because

$$\operatorname*{Res}_{z=0} \frac{\pi}{z^3 \sinh{(\pi z)} \sin{(\pi z)}} = \frac{\pi^3}{90}$$

the stated result follows. (NB that last residue at $z=0$ is best done using a Laurent series expansion, as the pole is of order 5.)

• (+1) The chances of you providing an insightful answer through complex analysis were very high. – Jack D'Aurizio Jan 9 '18 at 22:48
• @JackD'Aurizio: Glad to live up to expectations! – Ron Gordon Jan 9 '18 at 22:56

I realized my original proof can be shortened by exploiting symmetry.
The first three lines are unchanged: $$\frac{1}{\sinh z}=\frac{1}{z}+\sum_{m\geq 1}\left(\frac{1}{z-m\pi i}+\frac{1}{z+m\pi i}\right)(-1)^m$$ $$\frac{1}{\sinh(\pi n)}=\frac{1}{\pi n}+\frac{1}{\pi}\sum_{m\geq 1}\left(\frac{1}{n-mi}+\frac{1}{n+mi}\right)(-1)^m$$ $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)}=\frac{\eta(4)}{\pi}+\frac{2}{\pi}\sum_{m\geq 1}\sum_{n\geq 1}\frac{(-1)^{n+m+1}}{n^2(n^2+m^2)}$$ then by using $2\sum_{m,n\geq 1}f(m,n) = \sum_{m,n\geq 1}f(m,n)+f(n,m)$ we immediately get $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)}=\frac{\eta(4)-\eta(2)^2}{\pi}=\color{red}{\frac{\pi^3}{360}}$$ where $\eta(s)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}=(1-2^{1-s})\,\zeta(s)$ for any $s>1$.