Eisenstein sum.

Question

I have a proof of: $$S=\sum_{n=1}^{\infty}\frac{n(-1)^{n+1}}{(-1)^n+e^{\pi n}}=\frac{1}{24}.$$ That is related to Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$ . But the proof uses the relation involving the complete elliptic integral of the first kind and the Eisenstein series of weight 2. $$S=-\sum_{n=1}^{\infty}\frac{2n}{e^{2\pi n}+1}+\sum_{n=1}^{\infty}\frac{2n-1}{e^{(2n-1)\pi}-1},$$ where $$\sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}+1}=\frac{\Gamma{(1/4)}^4}{128\pi^3} -\frac{1}{24},$$ and $$\sum_{n=1}^{\infty}\frac{2n-1}{e^{(2n-1)\pi}-1}=\frac{\Gamma{(1/4)}^4}{64\pi^3} -\frac{1}{24}.$$ I'm looking for alternative proofs. Do you think that we can obtain this sum using Mellin tranform or using Contour Integration? Thanks for your cooperation.

Answer 1

As you noted the sum

$$S=\sum\limits_{n=1}^{\infty}\frac{n(-1)^{n+1}}{(-1)^n+e^{\pi n}}=\frac{1}{24}\tag{1}$$

can be written as

$$S=\sum\limits_{n=1}^{\infty} f(2n-1)+\sum\limits_{n=1}^{\infty} g(2n)\tag{2}$$

where

$$f(x)=\frac{x}{e^{\pi x}-1}\tag{3}$$

and

$$g(x)=-\frac{x}{e^{\pi x}+1}\tag{4}$$

with Mellin transforms

$$F(s)=\mathcal{M}_x\left[f(x)\right](s)=\int\limits_0^{\infty} \frac{x}{e^{\pi x}-1}\, x^{s-1} \, dx=\pi^{-s-1}\, \Gamma(s+1)\, \zeta(s+1)\tag{5}$$

and

$$G(s)=\mathcal{M}_x\left[g(x)\right](s)=\int\limits_0^{\infty } -\frac{x}{e^{\pi x}+1} x^{s-1} \, dx=\left(2^{-s}-1\right) \pi^{-s-1}\, \Gamma(s+1)\, \zeta(s+1)\tag{6}.$$

The inverse Mellin Transforms

$$f(x)=\mathcal{M}_s^{-1}[F(s)](x)=\frac{1}{2 \pi i} \int\limits_C F(s)\, x^{-s} ds\tag{7}$$

and

$$g(x)=\mathcal{M}_s^{-1}[G(s)](x)=\frac{1}{2 \pi i} \int\limits_C G(s)\, x^{-s}\, ds\tag{8}$$

lead to

$$S=\frac{1}{2 \pi i} \int\limits_C \left(F(s)\, \sum\limits_{n=1}^\infty (2n-1)^{-s}+G(s)\, \sum\limits_{n=1}^\infty (2n)^{-s}\right)\, ds$$ $$=\frac{1}{2 \pi i} \int\limits_C 4^{-s}\, \left(2^s-1\right)^2\, \pi^{-s-1}\, \Gamma(s+1)\, \zeta(s+1)\, \zeta(s)\, ds\tag{9}$$

where the integrand $4^{-s}\, \left(2^s-1\right)^2\, \pi^{-s-1}\, \Gamma(s+1)\, \zeta(s+1)\, \zeta(s)$ has a pole with a corresponding residue of $\frac{1}{24}$ at $s=\pm 1$, and I'm assuming the residue at $s=−1$ must be discarded for reasons similar to this answer to your related question.