Proof of convergence of series $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$

Question

I am trying to prove the convergence of $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}.$$ This is what I did: \begin{align} \sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n} &= \sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n} \cdot \frac{n+(-1)^n}{n+(-1)^n} \\&= \sum_{n=1}^{\infty}\frac{(-1)^n\cdot n + (-1)^{2n}}{n^2-(-1)^{2n}} \\&= \sum_{n=1}^{\infty}\frac{(-1)^n\cdot n + 1}{n^2-1} \\&= \sum_{n=1}^{\infty}\frac{(-1)^n\cdot n}{n^2-1} + \sum_{n=1}^{\infty}\frac{1}{n^2-1}. \end{align}

Now I am stuck, I don't know how to continue, because the first term doesn't even exists (dividing $0$). Can somebody help me?

Answer 1

Ignore the starting term and observe that $\sum \frac 1 {n^{2}+1}$ and $\sum (-1)^{n} \frac n {n^{2}-1}$ are both convergent series. [ Comparison with $\sum \frac 1 {n^{2}}$ for the first and alternating series test for the second].

Answer 2

Hint:

For even number of terms $$\sum_{r=1}^n\left(\dfrac{-1}{2r-1+1}+\dfrac1{2r-1}\right)=\sum_{r=1}^n\dfrac1{(2r-1)2r}$$

Use p-series convergence

For odd number of terms $$\sum_{r=1}^n\left(\dfrac{-1}{2r-1+1}+\dfrac1{2r-1}\right)-\dfrac1{2n}$$