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Consider the closed-form of the sum,

$$\sum_{n= 1}^\infty \frac{1}{n^p(e^{2\pi n}\pm1)} = \; ??$$

for $\color{blue}{p=4m+1}$. (Since closed-forms are known for $p=4m+3.$) For $p=1$, we have,

\begin{align} \sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}-1)} &=+\frac {3\log(\pi)}4-\log\Big(\Gamma\big(\tfrac 14\big)\Big)-\frac {\pi}{12}+\log(2)\\ \sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}+1)} &=-\frac{ 3\log(\pi)}4+\log\Big(\Gamma\big(\tfrac 14\big)\Big)\,+\,\frac {\pi}{4}-\frac{7\log(2)}4 \end{align}

and the related,

$$\log 2 = -\frac{4}{3}\sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}-1)} -\frac{4}{3}\sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}+1)}+\frac{2}{9}\pi$$

$$\zeta(5) = -\frac{72}{35}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}-1)} -\frac{2}{35}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}+1)}+\frac{1}{294}\pi^5$$

$$\zeta(9) = -\frac{992}{495}\sum_{n= 1}^\infty \frac{1}{n^9(e^{2\pi n}-1)} -\frac{2}{495}\sum_{n= 1}^\infty \frac{1}{n^9(e^{2\pi n}+1)}+\frac{125}{3704778}\pi^9$$

and so on, with the $\zeta(4m+1)$ formulas mentioned by Plouffe and Arndt in the 1998 article, "Identities Inspired from Ramanujan’s Notebooks". Note that by splitting the formula for $\log 2$, we get closed-forms for $\large{\sum_{n= 1}^\infty \frac{1}{n(e^{2\pi n}\pm1)}}$.

Question: Likewise, can we split the formula for $\zeta(5)$ and get a closed-form for,

$$\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}\pm1)} =\; ??$$

and other $p=4m+1,$ without expressing one case in terms of the other? Or without using other summations like,

\begin{align} \zeta(5) &= \frac{64}{5}\sum_{n= 1}^\infty \frac{1}{n^5(e^{\pi n}-1)}-\frac{74}{5}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}-1)}+\frac{1}{630}\pi^5\\[4pt] \zeta(5) &= \frac{256}{125}\sum_{n= 1}^\infty \frac{1}{n^5(e^{\pi n}+1)}-\frac{6}{125}\sum_{n= 1}^\infty \frac{1}{n^5(e^{2\pi n}+1)}+\frac{7}{2250}\pi^5 \end{align}

and similar variations?

P.S. In contrast, the concise closed-forms for $p=4m+3$ are already known for the negative case, examples in this post.

Tito Piezas III
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    The case for $p=1$ is simpler because there is a direct formula $f(q) =\sum_{n\geq 1}\frac{q^n}{n(1-q^n)}=-\log\prod_{n\geq 1}(1-q^n)$ so that it is essentially logarithm of Dedekind eta function. And one can notice via a little algebra that $\sum_{n\geq 1}\frac{q^n}{n(1+q^n)}=f(q)-2f(q^2)$. – Paramanand Singh Aug 25 '23 at 03:46
  • @ParamanandSingh I guess there is a qualitative difference between $\zeta(4m+1)$ and $\zeta(4m+3)$ since the former requires two Plouffe-type sums. Or maybe the closed-form involves an additional special function like in this post. C'mon, Singh, you found the closed-form for $p=4m+3$ and I know you can also do so for $p=4m+1$. :) – Tito Piezas III Aug 25 '23 at 05:12
  • I am using the Ramanujan Grosswald formula for odd zeta values, however I am not getting sums as mentioned in your post. I think there should be some relation between the functions $f_n(q) =\sum_{k\geq 1}\frac{k^{-n}q^n}{1-q^n}$ in the form of a differential equation. Such differential equations were given by Ramanujan for negative odd $n$ (essentially the differential equations connecting Eisenstein series of different weights). – Paramanand Singh Aug 26 '23 at 10:27

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