# Proof of 1eπ+1+3e3π+1+5e5π+1+…=124\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:

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

where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.

I then define for complex $\tau$ the function
so that

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

Now, let’s consider the function

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

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

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

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

Notes: 1)

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

For calculating the above limit, we used the facts

3) Here is the technique for computing the Mellin transform of $f(x)$.