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}

Question 1

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.

Question 2

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 $f(x)$ .

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}

Answer

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