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 n=1n oddnenπ+1=124.

I found a solution by myself 10 hours after I posted it, here it is:

f(x)=n=1n oddnxn1+xn,g(x)=n=1nxn1xn,

then I must prove that f(eπ)=124. It was not hard to find the relation between f(x) and g(x), namely f(x)=g(x)4g(x2)+4g(x4).

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 σ is the divisor function σ(n)=dnd.

I then define for complex τ the function
G2(τ)=π23(124n=1σ(n)e2πinτ) so that

But it is proven in Apostol “Modular forms and Dirichlet Series”, page 69-71 that G2(1τ)=τ2G2(τ)2πiτ, which gives {G2(i)=G2(i)+2πG2(i2)=4G2(2i)+4π. This is exactly was needed to get the desired result.

Hitoshigoto oshimai !

I find that sum fascinating. e,π 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 ζ(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 ζ(s) and the residue equals to 124. 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


= \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).

Source : Link , Question Author : danodare , Answer Author : Community

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