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I am looking for a proof of the following fabulous identity by Hardy and Ramanujan

$$\sum_{n = 1}^{\infty}\dfrac{1}{n^{2q - 1}}\left(a^{2q - 2}\coth\dfrac{n\pi b}{a} + (-1)^qb^{2q - 2}\coth\dfrac{n\pi a}{b}\right) \\ = \dfrac{2}{\pi ab}\sum_{k = 0}^{q}(-1)^{k - 1}\zeta(2k)\zeta(2q - 2k)a^{2q - 2k}b^{2k}$$

I'm most interested in a proof making use of complex analysis, Mellin transforms, Infinite series. However any approach is most welcomed.

Thanks.

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  • $\begingroup$ This is another reformulation of one of your earlier questions. It appears that question is deleted. The technique in this answer can be used to prove the identity in question. $\endgroup$
    – Paramanand Singh
    Apr 16, 2021 at 6:17
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    $\begingroup$ Also see a special case for $q=2$: math.stackexchange.com/q/907480/72031 $\endgroup$
    – Paramanand Singh
    Apr 16, 2021 at 6:20
  • $\begingroup$ Thanks, any idea on how to prove this one? $\endgroup$ Apr 16, 2021 at 6:40

1 Answer 1

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The proof below is borrowed from this paper by Bruce C. Berndt and Armin Straub.

Let us start with another identity of Ramanujan (proved here) : $$ \frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}=\frac{1}{2w}+\frac{1}{2}\log\frac{\beta}{\alpha}+\sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m^2\alpha} +\frac{m\beta\coth m\beta} {w-m^2\beta} \right) \tag{1}$$ which holds for positive $\alpha,\beta$ with $\alpha\beta=\pi^2$.

Next we can rewrite the right hand side of $(1)$ as a geometric series to get $$\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}=\frac{1}{2w}+\frac{1}{2}\log\frac{\beta}{\alpha}+\sum_{m=1}^{\infty} \left(\frac{1}{m}\sum_{k=0}^{\infty} \left(-\frac{w} {m^2\alpha}\right)^k\coth m\alpha- \frac{1}{m}\sum_{k=0}^{\infty} \left(\frac{w} {m^2\beta}\right)^k\coth m\beta \right) \tag{2}$$ And we can use the well known expansions of $\cot x, \coth x$ given by $$\cot x=\sum_{k=0}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!}x^{2k-1}\tag{3a}$$ and $$\coth x=\sum_{k=0}^{\infty} \frac{2^{2k}B_{2k}}{(2k)!}x^{2k-1}\tag{3b}$$ to rewrite left hand side of $(2)$ as $$\frac{\pi} {2}\left(\sum_{k=0}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!}(w\alpha)^{k-(1/2)}\right)\left(\sum_{j=0}^{\infty}\frac{2^{2j}B_{2j}}{(2j)!}(w\beta)^{j-(1/2)}\right)\tag{4}$$ The coefficient of $w^n$ in $(4)$ is $$\frac{\pi} {2}\cdot 2^{2n+2}\sum_{k=0}^{n+1}(-1)^{k}\frac {B_{2k}}{(2k)!}\cdot\frac{B_{2n+2-2k}}{(2n+2-2k)!}\alpha^{k-(1/2)}\beta^{n+(1/2)-k}$$ Since $\pi=\sqrt {\alpha\beta} $ we can rewrite the above as $$2^{2n+1}\sum_{k=0}^{n+1}(-1)^k\frac{B_{2k}}{(2k)!}\cdot\frac{B_{2n+2-2k}}{(2n+2-2k)!}\alpha^k\beta^{n+1-k}\tag{5}$$ Equating this with the coefficient of $w^n$ on right hand side of $(2)$ we get $$(-\alpha) ^{-n} \sum_{m=1}^{\infty} \frac{\coth m\alpha}{m^{2n+1}}-(\beta)^{-n}\sum_{m=1}^{\infty} \frac{\coth m\beta}{m^{2n+1}}= 2^{2n+1}\sum_{k=0}^{n+1}(-1)^k\frac{B_{2k}}{(2k)!}\cdot\frac{B_{2n+2-2k}}{(2n+2-2k)!}\alpha^k\beta^{n+1-k} $$ This is identical to your formula for $\alpha=\pi b/a, \beta=\pi a/b, q=n+1$ and note that your formula contains zeta values instead of Bernoulli numbers.


Ramanujan had a habit of analyzing partial fraction decomposition of various functions and deriving further identities by equating coefficients. What is really mysterious is his choice of functions like left hand side of $(1)$ on which to apply the partial fraction decomposition. And one can see that his technique avoids any complicated stuff.

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    $\begingroup$ The goddess maybe explains your last sentence. $\endgroup$
    – Erik Satie
    Apr 16, 2021 at 14:26
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    $\begingroup$ @ErikSatie: Yeah, Goddess Namagiri was the one he worshipped and Ramanujan attributed many of his discoveries to the goddess. $\endgroup$
    – Paramanand Singh
    Apr 16, 2021 at 15:03

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