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.
 A: Let's start with
$$
\sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1}
$$
Differentiating $(1)$ and multiplying by $x$, we get
$$
\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2}
$$
Taking the odd part of $(2)$ yields
$$
\sum_{n=0}^\infty(2n+1)x^{2n+1}=\frac{x(1+x^2)}{(1-x^2)^2}\tag{3}
$$
Using $(3)$, we get
$$
\begin{align}
\sum_{n=0}^\infty\frac{2n+1}{e^{(2n+1)\pi}+1}
&=\sum_{n=0}^\infty\sum_{k=1}^\infty(-1)^{k-1}(2n+1)e^{-(2n+1)k\pi}\\
&=\sum_{k=1}^\infty\sum_{n=0}^\infty(-1)^{k-1}(2n+1)e^{-(2n+1)k\pi}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\frac{e^{-k\pi}\left(1+e^{-2k\pi}\right)}{\left(1-e^{-2k\pi}\right)^2}\\
&=\frac12\sum_{k=1}^\infty(-1)^{k-1}\frac{\cosh(k\pi)}{\sinh^2(k\pi)}\tag{4}
\end{align}
$$

We can use the formula proven in this answer
$$
\pi\cot(\pi z)=\sum_{k\in\mathbb{Z}}\frac1{z+k}\tag{5}
$$
to derive
$$
\begin{align}
\pi\csc(\pi z)
&=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt]
&=\sum_{k\in\mathbb{Z}}\frac2{z+2k}-\sum_{k\in\mathbb{Z}}\frac1{z+k}\\
&=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}\\
\pi^2\frac{\cos(\pi z)}{\sin^2(\pi z)}
&=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(z+k)^2}\tag{6}
\end{align}
$$
then rotate coordinates with $z\mapsto iz$ to get
$$
\pi^2\frac{\cosh(\pi z)}{\sinh^2(\pi z)}=\sum_{j\in\mathbb{Z}}\frac{(-1)^j}{(z+ij)^2}\tag{7}
$$

Now plug $(7)$ into $(4)$:
$$
\begin{align}
\sum_{n=0}^\infty\frac{2n+1}{e^{(2n+1)\pi}+1}
&=\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j\in\mathbb{Z}}(-1)^{j+k-1}\frac1{(k+ij)^2}
\\
&=\frac1{2\pi^2}\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\\
&+\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j+k-1}\left(\frac1{(k+ij)^2}+\frac1{(k-ij)^2}\right)\\
&=\frac1{2\pi^2}\frac{\pi^2}{12}\\
&+\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j+k-1}\frac{2(k^2-j^2)}{(k^2+j^2)^2}\\
&=\frac1{24}+0\tag{8}
\end{align}
$$
A: $$S=\sum_{n=1}^{\infty }\frac{2n-1}{1+e^{\pi (2n-1)}},$$ 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 $x>0$
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 
$$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}}{1+t}dt+\int_{0}^{1}\frac{t^{x-1}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt\\
\\
\frac{\pi }{sin(\pi x)}=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-x}+\frac{1}{k-1+x})\\
\therefore from\ \frac{\pi }{sin(i\pi x)}=\int_{0}^{\infty }\frac{t^{-ix}+t^{ix-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-ix}+\frac{1}{k-1+ix})\\
\\
=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k+ix}{x^2+k^2}+\frac{k-1-ix}{(k-1^2)+x^2})=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k}{k^2+x^2}+\frac{k-1}{x^2+(k-1)^2})\\
\\
+i\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{x}{x^2+k^2}-\frac{x}{x^2+(k-1)^2})=i(\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}-\sum_{k=1}^{\infty }(-1)^{k}\frac{x}{x^2+k^2})\\
\\
\\
=i[\frac{-1}{x}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]=i[\frac{-1}{x}+2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]\\
\\\therefore \frac{\pi }{sinh(\pi x)}=\frac{1}{x}-2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{k^2+x^2}=\int_{0}^{1 }\frac{t^{-ix}+t^{ix-1}}{1+t}dt$$
A: The calculation of the Mellin transform of $f(x)$ is not present in the above answer, so I will show it here.
$$\mathfrak{M}\left(\frac{1}{e^{\pi x}+1};s \right) =
\int_0^\infty \frac{1}{e^{\pi x}+1} x^{s-1} dx =
\int_0^\infty \frac{1}{e^{\pi x}} \frac{1}{1+e^{-\pi x}} x^{s-1} dx \\=
\int_0^\infty \frac{1}{e^{\pi x}} \sum_{q\ge 0} (-1)^q e^{-\pi q x} x^{s-1} dx =
\int_0^\infty \sum_{q\ge 0} (-1)^q e^{-\pi (q+1) x} x^{s-1} dx \\
= \Gamma(s) \sum_{q\ge 0} (-1)^q \frac{1}{\pi^s (q+1)^s} 
= \frac{1}{\pi^s} \Gamma(s) (\zeta(s) - 2 \times 2^{-s} \times \zeta(s)) 
= \frac{1}{\pi^s} \Gamma(s) (1 - 2\times 2^{-s}) \zeta(s).$$
It now follows from the definition of the Mellin transform that
$$\mathfrak{M}\left(\frac{x}{e^{\pi x}+1};s \right) =
\frac{1}{\pi^{s+1}} \Gamma(s+1) (1 - 2^{-s}) \zeta(s+1).$$
A: 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 $f(x)$. 
A: Actually the above is not quite complete, the missing piece is the proof that we can drop the contribution from the pole at $s=-1,$ which is $x/24.$ To verify this we have to show that
$$\int_{-i\infty}^{i\infty} 
\frac{1}{\pi^{s+1}} \Gamma(s+1) (1-2^{-s})^2 \zeta(s+1)\zeta(s) ds = 0.$$
Now from the functional equation of the Riemann Zeta function we see that this integral is equal to
$$-\int_{-i\infty}^{i\infty}
\frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s})
\zeta(s) ds$$
Actually doing the accounting we find that the kernel 
$$ g(s) = \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s})
\zeta(s) $$ of this integral has the property that $g(s) = - g(-s)$ on the imaginary axis, so the integral is zero.
To see this consider what effect negation has on the individual terms.
$$\zeta(-s)\zeta(s) \to \zeta(s)\zeta(-s),$$
$$(2^s-1)(1-2^{-s}) \to (2^{-s}-1)(1-2^s) =  (2^s-1)(1-2^{-s}),$$
$$\sin(1/2 s\pi) \to \sin(1/2 (-s)\pi) = -\sin(1/2 s\pi).$$
The first two terms are even and the last one is odd, QED.
Note that we have taken advantage of the fact that $x=1$ ... for other values of $x$ this trick will not go through. Also relevant is that negation (rotation by 180 degrees about the origin) takes the imaginary axis to itself (this is not the case when we are integrating along some other line parallel to the imaginary axis in the right half plane).
A: 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
