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How to show $$\sum_{m=0}^{\infty} \sum _{n=0}^{\infty} \frac{e^{-2\pi(m+n)}}{m+n+1}=\frac{1}{2} (1+\coth \pi ).$$

I know that sum of IGP should work here, but how to disentangle m and n. Please help.

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Denote the rquired sum by $S$ and use $\frac{1}{x}=\int_{0}^{1} t^{x-1} dt.$ Then $$S=\sum_{m=0}^{\infty} \sum _{n=0}^{\infty} \frac{e^{-2\pi(m+n)}}{m+n+1} =\int_{0}^{1} \sum_{m=0}^{\infty} \sum _{n=0}^{\infty}~ (t e^{-2\pi})^{m+n} ~dt =\int_{0}^{1} \left(\sum_{k=0}^{\infty} (t e^{-2\pi})^k \right)^2 dt. $$ $$ \Rightarrow S=\int_{0}^{1} (1-te^{-2\pi})^{-2} dt .= \frac{1}{1-e^{-2\pi}}.$$

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Notice for any non-negative sequence $(f_{k})_{k\ge 0}$, we have $$\sum_{m=0}^\infty \sum_{n=0}^\infty f_{m+n} = \sum_{\ell=0}^\infty\sum_{n=0}^\ell f_{\ell} = \sum_{\ell=0}^\infty f_{\ell}\left(\sum_{n=0}^\ell 1\right) = \sum_{\ell=0}^\infty (\ell+1)f_{\ell}$$ Since the sequence is non-negative, it is legal to perform the sum in any order.

For this particular problem, substitute $f_k$ by $\frac{e^{-2\pi k}}{k+1}$ will turn your sum into a geometric series.

$$\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{e^{-2\pi(m+n)}}{m+n+1} = \sum_{\ell=0}^\infty e^{-2\pi \ell} = \frac{1}{1 - e^{-2\pi}} = \frac{e^{\pi}}{2\sinh(\pi)}\\ = \frac{\cosh(\pi)+\sinh(\pi)}{2\sinh(\pi)} = \frac12(\coth(\pi) +1) $$

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