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

g(x)=n=1σ(n)xn

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
f(eπ)=g(eπ)4g(e2π)+4g(e4π)=124+G2(i2)+4G2(i)4G2(2i)8π2.

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.

Answer

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

F(s)=0xs1f(x)dx,f(x)=12πic+icixsF(s)ds.

Now, let’s consider the function

f(x)=xeπx+1.

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

F(s)=πs1Γ(s+1)(12s)ζ(s+1),

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

xeπx+1=12πiCπs1Γ(s+1)(12s)ζ(s+1)xsds.

Substituting x=2n+1 and summing yields

n=02n+1eπ(2n+1)+1=12πiCπs1Γ(s+1)(12s)ζ(s+1)n=0(2n+1)sds

=12πiCπs1Γ(s+1)(12s)2ζ(s+1)ζ(s)ds.

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

n=02n+1eπ(2n+1)+1=124

Notes: 1)

n=0(2n+1)s=(12s)ζ(s).

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

r=lims=1(s1)(πs1Γ(s+1)(2s1)2ζ(s+1)ζ(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).

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

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