Intereting Posts

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$$-\frac{8\varepsilon_0V_0}{\pi}\sum_{n \text{ odd}}\frac{1}{n\sinh(n\pi)}=\boxed{\displaystyle-\frac{\varepsilon_0V_0}{\pi}\ln 2.}$$

I have not found a way to sum this series analytically. Mathematica gives the numerical value $0.0866434$, which agrees precisely with $\ln2/8.$

Can someone do this series please?

it is from the book on electrodynamics by Griffiths.

Its in the solution manual 3.48.

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- Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$
- Clarification on how to prove polynomial representations exist for infinite series

**Lemma**. For $x>0$, we have

$$\sum_{n\text{ odd}}\frac{1}{n\sinh nx}=-\frac12\,\ln\frac{\vartheta_4\left(0|e^{-x}\right)}{\vartheta_3\left(0|e^{-x}\right)}.\tag{1}$$

**Proof**. Let us write

$$\frac{1}{\sinh nx}=\frac{2}{e^{nx}}\sum_{k=0}^{\infty}e^{-2knx}=

2\sum_{k=0}^{\infty}e^{-(2k+1)nx}.$$

Substituting this into the left side of (1) and using that $\displaystyle\sum_{n\text{ odd}}\frac{q^n}{n}=-\frac12\ln\frac{1-q}{1+q}$, we get

$$\sum_{n\text{ odd}}\frac{1}{n\sinh nx}=-\ln\prod_{k=0}^{\infty}\frac{1-e^{-(2k+1)x}}{1+e^{-(2k+1)x}}=-\frac12\,\ln\frac{\vartheta_4\left(0|e^{-x}\right)}{\vartheta_3\left(0|e^{-x}\right)},$$

where at the last step we have used product representations of the Jacobi theta functions (see e.g. (92), (93) here). $\blacksquare$

The formula

$$\sum_{n\text{ odd}}\frac{1}{n\sinh \pi n}=\frac{\ln2}{8}$$

then immediately follows from the special values

\begin{align*}

\vartheta_3\left(0|e^{-\pi}\right)=\frac{\pi^{\frac14}}{\Gamma\left(\frac34\right)},\qquad

\vartheta_4\left(0|e^{-\pi}\right)=\frac{\pi^{\frac14}}{2^{\frac14}\Gamma\left(\frac34\right)}.

\end{align*}

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