Is :$\sum_{n=2}^{\infty}\frac{(-1)^n}{1-e^n}=0.5\sum_{n=2}^{\infty}\frac{1}{1-e^n}$?

Question

$$S_1=\sum_{n=2}^{\infty}\frac{(-1)^n}{1-e^n}=-0.11\cdots \text{ and }\ S_2=\sum_{n=2}^{\infty}\frac{1}{1-e^n}=-0.23\cdots$$ on Wolfram Alpha: (first sum and second sum).

I'm unable to show if I'm true that $S_1=0.5 S_2$ because I can't come up to the closed form of both $S_1$ and $S_2$

Answer 1

Using Mathematica can be found closed form solution:

$$\sum _{n=2}^{\infty } \frac{(-1)^n}{1-\exp (n)}=\ln (-1+e)+\psi _e(-i \pi )-\frac{1}{-1+e}+i \pi$$

$$\sum _{n=2}^{\infty } \frac{1}{1-\exp (n)}=-1-\frac{1}{1-e}+\ln (-1+e)+\psi _{\frac{1}{e}}(1)$$

where: $\psi _{\frac{1}{e}}(1)$ and $\psi _e(-i \pi )$ is q-digamma function

EDITED:

Alternative to WolframAlfa to calculate numerically sum.

Go to SymPy web page and paste code:Sum(1/(1-exp(n)), (n, 2, oo)).evalf(50)

and click Evaluate Button.