<s>[INST] Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$ I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$. 
I found a solution by myself 10 hours after I posted it, here it is:
$$f(x)=\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{nx^n}{1+x^n},\quad\quad g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n},$$
then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.
Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get
$$g(x)=\sum_{n=1}^{\infty}\sigma(n)x^n$$
where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.
I then define for complex $\tau$ the function 
$$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that
$$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$
But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.
Hitoshigoto oshimai !
I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!
Thanks to everyone who contributed.
 [/INST] Integrate the function $$f(z) =   \frac{z e^{iz}}{\cosh (z) + \cos(z)} \, $$ around the contour $[-\sqrt{2} \pi N, \sqrt{2} \pi N] \cup \sqrt{2} \pi Ne^{i[0, \pi]}$, where $N$ is a positive integer.
The integral vanishes on the semicircle as $N \to \infty$ because $|f(z)|$ decays exponentially fast to zero along the entire semicircle.
We therefore have
$\begin{align} \int_{-\infty}^{\infty}f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (1+i)}{2} \right] + \sum_{n=0}^{\infty}\operatorname{Res} \left[f(z),  \frac{(2n+1) \pi (-1+i)}{2} \right]\right)  \\ &= 2 \pi i \left(\sum_{n=0}^{\infty} \lim_{z \to  \frac{(2n+1) \pi (1+i)}{2}  }\frac{ze^{iz}}{\sinh(z) - \sin(z)} + \sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (-1+i)}{2}  }\frac{ze^{iz}}{\sinh(z) - \sin(z)} \right) \\ &= 2 \pi i \left(\frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)} + \frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)}\right) \\ &= 2 \pi^{2} i  \sum_{n=0}^{\infty} \frac{(2n+1) e^{- (2n+1) \pi /2}}{\cosh \left(\frac{(2n+1)\pi }{2} \right)}  \\ &= 4 \pi^{2}i \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1)\pi }+1}. \end{align}$
Equating the imaginary parts on both sides of the equation, we get $$  \begin{align} \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1) \pi }+1} &= \frac{1}{4 \pi^{2}} \int_{-\infty}^{\infty} f(x) \, \mathrm dx \\ &= \frac{1}{2 \pi^{2}}\int_{0}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)} \, \mathrm dx  \\ &= \frac{1}{\pi^{2}} \, \Im \int_{0}^{\infty} x \sum_{n=1}^{\infty} (-1)^{n-1} e^{(-1+i)nx} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \sum_{n=1}^{\infty} (-1)^{n-1}  \int_{0}^{\infty} x e^{(-1+i)nx} \, \mathrm dx \\ &=  \frac{1}{\pi^{2}} \, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{(1-i)^{2}n^{2}} \\ &= \frac{1}{2 \pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}  \\ &=  \frac{1}{24} . \end{align} $$

In general, integrating the function $$g(z) =  \frac{z^{4k+1} e^{iz}}{\cosh (z) + \cos(z)} \, , \quad k \in \mathbb{N}_{\ge 0},$$ around the same contour shows that
$$ \begin{align} \sum_{n=0}^{\infty} \frac{(2n+1)^{4k+1}}{e^{(2n+1)\pi}+1} &= \frac{(-1)^{k}2^{2k-1}}{\pi^{4k+2}} \int_{0}^{\infty} \frac{x^{4k+1} \sin(x)}{\cosh(x) + \cos(x)} \, \mathrm dx \\ &= \frac{(-1)^{k}2^{2k-1}}{\pi^{4k+2}} \frac{1}{2^{2k}} (-1)^{k} (4k+1)! \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{4k+2}} \\ &\overset{(1)}{=} \frac{B_{4k+2}}{2 (4k+2)}  \left(2^{4k+1}-1\right), \end{align}$$ where $B_{k}$ is the $k$th Bernoulli number.

$(1)$ https://en.wikipedia.org/wiki/Dirichlet_eta_function#Particular_values
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