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