This sum is from Ramanujan's letters to G. H. Hardy and Ramanujan gives the summation formula as \begin{align} &\frac{1}{1^{3}}\left(\coth \pi x + x^{2}\coth\frac{\pi}{x}\right) + \frac{1}{2^{3}}\left(\coth 2\pi x + x^{2}\coth\frac{2\pi}{x}\right) \notag\\ &\, \, \, \, \, \, \, \, + \frac{1}{3^{3}}\left(\coth 3\pi x + x^{2}\coth\frac{3\pi}{x}\right) + \cdots\notag\\ &\, \, \, \, \, \, \, \, = \frac{\pi^{3}}{90x}(x^{4} + 5x^{2} + 1)\notag \end{align} Since $$\coth x = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} = \frac{1 + e^{-2x}}{1 - e^{-2x}} = 1 + 2\frac{e^{-2x}}{1 - e^{-2x}}$$the above sum is transformed into $$(1 + x^{2})\sum_{n = 1}^{\infty}\frac{1}{n^{3}} + 2\sum_{n = 1}^{\infty}\frac{e^{-2n\pi x}}{n^{3}(1 - e^{-2n\pi x})} + 2x^{2}\sum_{n = 1}^{\infty}\frac{e^{-2n\pi/x}}{n^{3}(1 - e^{-2n\pi/x})}$$ If we put $q = e^{-\pi x}$ we get sums like $\sum q^{2n}/\{n^{3}(1 - q^{2n})\}$ which I don't know how to sum.

It seems I am going on a wrong track. Please provide some alternative approach.

Update: All the answers given below so far use complex analyis (transforms and residues) to evaluate the sum. I am almost certain that Ramanujan did not evaluate the sum using complex analysis. Perhaps the method by Ramanujan is more like the one explained in this question. Do we have any approach based on real-analysis only?

  • 1
    $\begingroup$ The series you get at the end look like Lambert series, which have known resummation properties. But there's not an obvious multiplicative function in there. $\endgroup$ Sep 27, 2014 at 4:50
  • $\begingroup$ @Semiclassical: I am aware of some standard results in Lambert series, but Ramanujan's results are so non-obvious in general that I am really unable to proceed here using the Lambert series. $\endgroup$
    – Paramanand Singh
    Sep 27, 2014 at 5:01
  • $\begingroup$ Something odd: If you take $x\mapsto x^{-1}$, then the LHS of the initial identity essentially picks up a factor of $1/x^2$ but the RHS picks up a factor of $x^2$. $\endgroup$ Sep 27, 2014 at 5:46
  • $\begingroup$ @Semiclassical: you are right. Maybe its a typo and I need to replace $90x^3$ by $90x$. I will double check and edit. $\endgroup$
    – Paramanand Singh
    Sep 27, 2014 at 5:52
  • $\begingroup$ @Semiclassical: I have obtained this problem from collected papers of Ramanujan and the same typo is there too. I will correct it here. $\endgroup$
    – Paramanand Singh
    Sep 27, 2014 at 5:57

4 Answers 4


It seems to have escaped attention that this sum may be evaluated using harmonic summation techniques.

Put $$S(x) = \zeta(3) + \sum_{n\ge 1} \frac{-1+\coth(n\pi x)}{n^3}$$ and introduce the sum $$T(x) = \sum_{n\ge 1} \frac{-1+\coth(n\pi x)}{n^3}.$$

The sum term is harmonic and may be evaluated by inverting its Mellin transform. We will construct a functional equation for $T(x).$

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 $$\lambda_k = \frac{1}{k^3}, \quad \mu_k = k \quad \text{and} \quad g(x) = 2\frac{e^{-2\pi x}}{1-e^{-2\pi x}}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$2 \int_0^\infty \frac{e^{-2\pi x}}{1-e^{-2\pi x}} x^{s-1} dx \\ = 2 \int_0^\infty \sum_{q\ge 1} e^{-2q\pi x} x^{s-1} dx = 2 \sum_{q\ge 1} \int_0^\infty e^{-2q\pi x} x^{s-1} dx \\= 2 \Gamma(s) \sum_{q\ge 1} \frac{1}{(2\pi q)^s} = \frac{2}{2^s} \frac{1}{\pi^s} \Gamma(s) \zeta(s).$$

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

$$Q(s) = \frac{2}{2^s} \frac{1}{\pi^s} \Gamma(s) \zeta(s) \zeta(s+3) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^3} \frac{1}{k^s} = \zeta(s+3)$$ for $\Re(s) > -2.$

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 $\Re(s) = -3 -1/2$ we get $$T(x) = \frac{\pi^3 x^3}{90} + 4\zeta'(-2)\pi^2 x^2 + \frac{\pi^3 x}{18} - \zeta(3) + \frac{\pi^3}{90x} + \frac{1}{2\pi i} \int_{-7/2-i\infty}^{-7/2+i\infty} Q(s)/x^s ds.$$

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

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 $-Q(t).$

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

We have established the functional equation $$T(x) = \frac{\pi^3 x^3}{90} + 4\zeta'(-2)\pi^2 x^2 + \frac{\pi^3 x}{18} - \zeta(3) + \frac{\pi^3}{90x} - x^2 T(1/x).$$

Finally returning to the sum that was the initial goal we see that it has the value $$\zeta(3) + T(x) + x^2 (\zeta(3) + T(1/x))$$ or $$\zeta(3) + T(x) + x^2 \zeta(3) + x^2 T(1/x).$$ Using the functional equation for $T(x)$ this becomes $$\zeta(3) + T(x) + x^2 \zeta(3) + \frac{\pi^3 x^3}{90} + 4\zeta'(-2)\pi^2 x^2 + \frac{\pi^3 x}{18} - \zeta(3) + \frac{\pi^3}{90x} - T(x)$$ which is $$x^2 \zeta(3) + \frac{\pi^3 x^3}{90} + 4\zeta'(-2)\pi^2 x^2 + \frac{\pi^3 x}{18} + \frac{\pi^3}{90x}.$$

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

Addendum. In view of the fact that $$\zeta(3) + 4\zeta'(-2)\pi^2 =0 $$ (consult e.g. MathWorld) this finally becomes $$\frac{\pi^3 x^3}{90} + \frac{\pi^3 x}{18} + \frac{\pi^3}{90x} = \frac{\pi^3}{90x} \left(x^4 + 5x^2 + 1\right).$$

Addendum II. There is another functional equation of a harmonic sum at this MSE link, this one somewhat more advanced.

  • $\begingroup$ Interesting solution and interesting reference (+1) $\endgroup$ Sep 30, 2014 at 6:26
  • $\begingroup$ Do we have any approach based on real-analysis only? See the update in my question. $\endgroup$
    – Paramanand Singh
    Oct 1, 2014 at 8:17
  • 1
    $\begingroup$ A theorem that I became aware of through posts on MSE is the Ramanujan Master Theorem. This would seem to indicate a profound understanding of Mellin transform methods. $\endgroup$ Oct 1, 2014 at 22:47
  • $\begingroup$ @MarkoRiedel: It appears that although Ramanujan does not use complex analysis explicitly, but has deep conception of some powerful results which can be obtained via complex analytic methods. Seen from that perspective perhaps your answer is closer to what Ramanujan might have done. I will accept it. $\endgroup$
    – Paramanand Singh
    Oct 2, 2014 at 5:50

Recall the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote $\mathbb{W}=\mathbb{Z}/\{0\}$) :

$$\sum_{m\in\mathbb{W}}\frac{1}{m^2+z^2}=\frac{\pi\coth\pi z}{z}-\frac{1}{z^2}\tag{ML}$$

Hence, your sum is by its symmetry :

$$\begin{align} S&=\frac{1}{2}\sum_{n \in \mathbb{W}}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3} \\ \\ &=\frac{1}{2\pi x}\sum_{n \in \mathbb{W}}\left(\frac{1}{n^4}+\sum_{m\in\mathbb{W}}\frac{x^2/n^2}{m^2+n^2x^2}\right) +\left(\frac{x^4}{n^4}+\sum_{m\in\mathbb{W}}\frac{x^2/n^2}{m^2+n^2/x^2}\right)\tag{1}\\ \\ &=\frac{1}{2\pi x}\left(\zeta(4)+x^4\zeta(4)+\sum_{n,m \in \mathbb{W}^2}\frac{x^2}{n^2}\frac{1}{m^2+n^2x^2}+\frac{x^2}{n^2}\frac{1}{m^2+n^2/x^2}\right)\tag{2}\\ \\ &=\frac{1}{2\pi x}\left(2\zeta(4)+2x^4\zeta(4)+x^2\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2m^2}\frac{m^2+n^2x^2}{m^2+n^2x^2}\right)\tag{3}\\ \\ &=\frac{1}{2\pi x}\left(2\zeta(4)+2x^4\zeta(4)+4x^2\zeta^2(2)\right)\tag{4}\\ \\ &=\frac{1}{2\pi x}\left(2\frac{\pi^4}{90}+2x^4\frac{\pi^4}{90}+4x^2\frac{\pi^4}{36}\right)\tag{5}\\ \\ &=\frac{\pi^3}{90x}\left(1+x^4+5x^2\right) \end{align}$$


$(1)$ Use the Mittag-Leffler formula (ML) with $z=nx$ and $z=n/x$

$(2,4)$ Recall $\zeta(s)=\sum_{n=1}^{\infty}1/n^s$

$(3)$ In the second sum rename $n \longleftrightarrow m$

$(5)$ Zetas for $s=2$ and $4$ are well known, i.e. $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$

  • 1
    $\begingroup$ This one appears to be simplest and very direct. +1 $\endgroup$
    – Paramanand Singh
    Jul 22, 2016 at 16:04
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    $\begingroup$ @ParamanandSingh At least it is a method not involving complex analysis at all since even ML formula could be obtained by using noncomplex methods, for example performing fourier expansion of the function $f(x)=\cosh ax$ on the interval $x\in\left[-\pi,\pi\right]$, as in here. By the way by the same argument is $\sum_{n=1}^{\infty}\frac{\sinh\big(\pi n\sqrt2\big)-\sin\big(\pi n\sqrt2\big)}{n^3\Big({\cosh\big(\pi n\sqrt2}\big)-\cos\big(\pi n\sqrt2\big)\Big)}$ found in closed form. $\endgroup$
    – Machinato
    Jul 23, 2016 at 20:36
  • $\begingroup$ @Machinato Nice derivation, is your proof unique or did you read this somewhere? $\endgroup$ Apr 27, 2021 at 3:22
  • 2
    $\begingroup$ @BooleanWick I came with it on my own. But with almost everything in math, it is possible that somebody else has found this proof as well. $\endgroup$
    – Machinato
    Apr 28, 2021 at 7:31

Following in the same manner as this answer...

We are going to use the contour integral $$ \oint\pi\cot\left(\frac{\pi z}{\pi x}\right)\left(\frac{\coth(z)}{z^3}-\frac1{z^4}-\frac1{3z^2}\right)\mathrm{d}z=0\tag{1} $$ where the contours of interest are, for real $R\to\infty$ and integer $n\to\infty$, $$ \small\textstyle\color{#00A000}{[R,-R]+(n+\frac12)\pi i}\cup\color{#C00000}{-R+(n+\frac12)\pi i[1,-1]}\cup\color{#00A000}{[-R,R]-(n+\frac12)\pi i}\cup\color{#C00000}{R+(n+\frac12)\pi i[-1,1]} $$ The integral along the red paths becomes negligible as $R\to\infty$. Along the upper green path, where $\mathrm{Im}(z)\approx+\infty$, $\cot(z)\approx-i$. Along the lower green path, where $\mathrm{Im}(z)\approx-\infty$, $\cot(z)\approx+i$. Since $\coth(z+\frac\pi2i)=\tanh(z)$, the integral along each of the green paths tends to $0$. Therefore, the full integral is $0$.

Since $$ \pi\cot\left(\frac{\pi z}{\pi x}\right)\text{ has residue }\pi x\text{ at }z=\pi nx\tag{2} $$ and $$ \frac{\coth(z)}{z^3}-\frac1{z^4}-\frac1{3z^2}=-\frac1{45}+O(z^2)\text{ at }z=0\tag{3} $$ the contribution from the singularities on the real axis is $$ 2\pi i\cdot\pi x\left[2\sum_{n=1}^\infty\left(\frac{\coth(\pi nx)}{(\pi nx)^3}-\frac1{(\pi nx)^4}-\frac1{3(\pi nx)^2}\right)-\frac1{45}\right]\tag{4} $$ Since $$ \frac{\coth(z)}{z^3}\text{ has residue }\frac1{(\pi in)^3}\text{ at }z=\pi i n\text{ for }n\ne0\tag{5} $$ and $$ \pi\cot\left(\frac{\pi z}{\pi x}\right)=-\pi i\coth\left(\frac{\pi n}{x}\right)\text{ at }z=\pi in\tag{6} $$ the contribution from the singularities on the imaginary axis is $$ 2\pi i\left[2\sum_{n=1}^\infty\frac\pi{x^3}\frac{\coth\left(\frac{\pi n}{x}\right)}{\left(\frac{\pi n}{x}\right)^3}\right]\tag{7} $$ Combining $(1)$, $(4)$, and $(7)$, yields $$ x^2\sum_{n=1}^\infty\frac{\coth(\pi nx)}{(\pi nx)^3}+\frac1{x^2}\sum_{n=1}^\infty\frac{\coth\left(\frac{\pi n}{x}\right)}{\left(\frac{\pi n}{x}\right)^3} =\frac{\zeta(4)}{\pi^4x^2}+\frac{\zeta(2)}{3\pi^2}+\frac{x^2}{90}\tag{8} $$ Multiplying by $\pi^3x$ to match the question, we get $$ \sum_{n=1}^\infty\frac{\coth(\pi nx)+x^2\coth(\pi n/x)}{n^3}=\frac{\pi^3}{90x}\left(1+5x^2+x^4\right)\tag{9} $$

  • $\begingroup$ Seems to be a good day for me to see interesting solutions! (+1) $\endgroup$ Sep 30, 2014 at 14:17
  • $\begingroup$ Thanks robjohn. This seems simpler than the other answer from Marko. +1 $\endgroup$
    – Paramanand Singh
    Sep 30, 2014 at 16:17
  • $\begingroup$ Do we have any approach based on real-analysis only? See the update in my question. $\endgroup$
    – Paramanand Singh
    Oct 1, 2014 at 8:18

Yet another approach using contour integration is to integrate the function $$f(z) = \frac{\pi \cot (\pi z) \coth (\pi x z)}{z^{3}} $$ around a circle centered at the origin that avoids the poles on the real and imaginary axes.

If we let the radius of the circle go to infinity discretely, the integral will vanish.

So summing the residues, we get $$2 \sum_{n=1}^{\infty} \frac{\coth (n \pi x)}{n^{3}} + \sum_{n=1}^{\infty} \frac{\cot (\frac{in \pi}{x})}{x(\frac{in}{x})^{3}} + \sum_{n=1}^{\infty} \frac{\cot (-\frac{i n \pi}{x})}{x (-\frac{in}{x} )^{3}} + \text{Res}[f(z),0] = 0,$$

which implies

$$\sum_{n=1}^{\infty} \frac{\coth (n \pi x)}{n^{3}} + x^{2} \sum_{n=1}^{\infty} \frac{\coth(\frac{n \pi}{x})}{n^{3}} = - \frac{1}{2} \ \text{Res} [f(z),0]. $$

Expanding at the origin, we get

$$ \begin{align} \small f(z) &= \frac{\pi}{z^{3}}\left(\frac{1}{\pi z} - \frac{2 \zeta(2)}{\pi} z-\frac{2 \zeta(4)}{\pi} z^{3} + \mathcal{O}(z^{5})\right) \left(\frac{1}{\pi (xz)} + \frac{2 \zeta(2)}{\pi} (xz) - \frac{2 \zeta(4)}{\pi} (xz)^{3} + \mathcal{O}(z^{5}) \right) \\ &= \frac{1}{\pi x} \frac{1}{z^{5}} + \frac{2 \zeta(2) x^{2}-2 \zeta(2)}{\pi x} \frac{1}{z^{3}} {\color{red}{-\frac{2 \zeta(4) x^{4}+4 \zeta(2)^{2} x^{2} + 2 \zeta(4)}{\pi x}}} \frac{1}{z} + \mathcal{O}(z) .\end{align} $$

Therefore, $$ \sum_{n=1}^{\infty} \frac{\coth (n \pi x)}{n^{3}} + x^{2} \sum_{n=1}^{\infty} \frac{\coth(\frac{n \pi}{x})}{n^{3}} = \frac{\zeta(4) x^{4}+2 \zeta(2)^{2} x^{2} + \zeta(4)}{\pi x} =\frac{\pi^{3}}{90x} \left( x^{4}+5x^{2}+1 \right). $$

  • $\begingroup$ Seems like I am getting simpler and simpler answers. +1 $\endgroup$
    – Paramanand Singh
    Oct 1, 2014 at 8:19
  • $\begingroup$ Do we have any approach based on real-analysis only? See the update in my question. $\endgroup$
    – Paramanand Singh
    Oct 1, 2014 at 8:20
  • $\begingroup$ This works the same as my answer, except that $(3)$ and $(4)$ take the place of your expansion at the origin. (+1) $\endgroup$
    – robjohn
    Oct 1, 2014 at 8:26

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