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.
We will use the Mellin transform technique. Recalling the Mellin transform and its inverse


  $$ F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds.  $$


Now, let's consider the function


  $$ f(x)= \frac{x}{e^{\pi x}+1}. $$


Taking the Mellin transform of $f(x)$, we get


  $$ F(s)={\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left(1- {2}^{-s} \right) 
\zeta  \left( s+1 \right),$$


where $\zeta(s)$ is the zeta function . Representing the function in terms of the inverse Mellin Transform, we have


  $$ \frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left( 1-{2}^{-s} \right) 
\zeta  \left( s+1 \right) x^{-s}ds. $$


Substituting $x=2n+1$ and summing yields

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma  \left( s+1 \right)\left(1-{2}^{-s} \right) 
\zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$

$$ = \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$

Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by

$$ \sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24} $$

Notes: 1)  


  $$ \sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta  \left( s \right).  $$


2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as

$$ r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$

$$  =  \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1}  {\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right)  = \frac{1}{24}. $$

For calculating the above limit, we used the facts

$$ \lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}. $$

3) Here is the technique for computing the Mellin transform of $f(x)$.
$$S=\sum_{n=1}^{\infty }\frac{2n-1}{1+e^{\pi (2n-1)}},$$ first we know that $$\frac{1}{1+e^x}=\sum_{k=1}^{\infty }(-1)^{k-1}e^{-kx}$$
therefore 
$$S=\sum_{n=1}^{\infty }(2n-1)\sum_{k=1}^{\infty }(-1)^{k-1}e^{-k(2n-1)\pi }=\sum_{k=1}^{\infty }(-1)^{k-1}\sum_{n=1}^{\infty }(2n-1)e^{-k(2n-1)\pi }$$

but we know
$$ \sum_{n=1}^{\infty }e^{-(2n-1)\pi x}=\frac{1}{2sinh(\pi x)},$$ 
for all $x&gt;0$
but
 $$\frac{1}{2sinh(\pi x)}=(\frac{i}{2\pi })(\frac{\pi }{sin(i\pi x)})$$ and  $$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\frac{\pi }{sin(\pi x)}$$ 
therefore 
$$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}}{1+t}dt+\int_{0}^{1}\frac{t^{x-1}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt\\
\\
\frac{\pi }{sin(\pi x)}=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-x}+\frac{1}{k-1+x})\\
\therefore from\ \frac{\pi }{sin(i\pi x)}=\int_{0}^{\infty }\frac{t^{-ix}+t^{ix-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-ix}+\frac{1}{k-1+ix})\\
\\
=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k+ix}{x^2+k^2}+\frac{k-1-ix}{(k-1^2)+x^2})=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k}{k^2+x^2}+\frac{k-1}{x^2+(k-1)^2})\\
\\
+i\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{x}{x^2+k^2}-\frac{x}{x^2+(k-1)^2})=i(\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}-\sum_{k=1}^{\infty }(-1)^{k}\frac{x}{x^2+k^2})\\
\\
\\
=i[\frac{-1}{x}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]=i[\frac{-1}{x}+2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]\\
\\\therefore \frac{\pi }{sinh(\pi x)}=\frac{1}{x}-2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{k^2+x^2}=\int_{0}^{1 }\frac{t^{-ix}+t^{ix-1}}{1+t}dt$$