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=1nxn1−xn,

then I must prove that f(e−π)=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)=∑d∣nd.

I then define for complex τ the function

G2(τ)=π23(1−24∞∑n=1σ(n)e2πinτ) so that

f(e−π)=g(e−π)−4g(e−2π)+4g(e−4π)=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)=∫∞0xs−1f(x)dx,f(x)=12πi∫c+i∞c−i∞x−sF(s)ds.

Now, let’s consider the function

f(x)=xeπx+1.

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

F(s)=π−s−1Γ(s+1)(1−2−s)ζ(s+1),

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

xeπx+1=12πi∫Cπ−s−1Γ(s+1)(1−2−s)ζ(s+1)x−sds.

Substituting x=2n+1 and summing yields

∞∑n=02n+1eπ(2n+1)+1=12πi∫Cπ−s−1Γ(s+1)(1−2−s)ζ(s+1)∞∑n=0(2n+1)−sds

=12πi∫Cπ−s−1Γ(s+1)(1−2−s)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=(1−2−s)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(s−1)(π−s−1Γ(s+1)(2−s−1)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*