Does the series of $\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}$ converge? Does the following sum converge?
$$
  \sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}
$$
Have tried ratio test and root test, inconclusive. alternating series test not useful.

If $\sum_{n=2}^{\infty}(-1)^n \frac{n+1}{n-1}$ converge, then there exist a $N$ such that $\forall n \geq N$
$$L-\frac{1}{3} < \sum_{n=2}^{2N}(-1)^n \frac{n+1}{n-1} < L+\frac{1}{3}$$
But then 
$$L-\frac{1}{3}-\frac{2N+2}{2N} < \sum_{n=2}^{2N+1}(-1)^n \frac{n+1}{n-1} < L+\frac{1}{3} - \frac{2N+2}{2N}$$
But $\frac{2N+2}{2N}> \frac{2}{3}$, then
$$L-1 < \sum_{n=2}^{2N+1}(-1)^n \frac{n+1}{n-1} < L-\frac{1}{3} $$
contradiction
This is the idea of the proof of why the general term must converge to 0
