find sum of $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$

Question

I want to find the exact sum of this expression: $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$

I've already proved that it converges by condition. Also, I think that it's sort of telescoping series, because if I open it I get: $\frac{-1}{2}+\frac{1}{1}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\frac{1}{5}...$

But I cant think of way to calculate it.

Answer 1

Hint. Note that if $N$ is even then $$\begin{align}S_N&=\sum_{n=1}^{N}\frac{(-1)^n}{n-(-1)^n}= -\frac{1}{2}+\frac{1}{1}-\frac{1}{4}+\frac{1}{3}+\dots-\frac{1}{N}+\frac{1}{N-1}\\&= \frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{N-1}-\frac{1}{N}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}, \end{align}$$ and $$S_{N+1}=S_N+\frac{(-1)^{N+1}}{N+1-(-1)^{N+1}}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n} -\frac{1}{N+2}.$$