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.
 A: Hint:
What is
$$\lim\limits_{n\to\infty}\frac{n+1}{n-1}? $$
A: 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
A: Let
$s(m)
=\sum_{n=2}^{m} \frac{(-1)^n(n+1)}{n-1}
$.
$\begin{array}\\
s(2m+1)
&=\sum_{n=2}^{2m+1} \frac{(-1)^n(n+1)}{n-1}\\
&=\sum_{n=1}^{m} (\frac{(-1)^{2n}(2n+1)}{2n-1}+\frac{(-1)^{2n+1}(2n+1+1)}{2n+1-1})\\
&=\sum_{n=1}^{m} (\frac{(2n+1)}{2n-1}-\frac{(2n+2)}{2n})\\
&=\sum_{n=1}^{m} (\frac{2n-1+2}{2n-1}-\frac{n+1}{n})\\
&=\sum_{n=1}^{m} (1+\frac{2}{2n-1}-(1+\frac{1}{n}))\\
&=\sum_{n=1}^{m} (\frac{2}{2n-1}-\frac{1}{n})\\
&=\sum_{n=1}^{m} \frac{2n-(2n-1)}{n(2n-1)}\\
&=\sum_{n=1}^{m} \frac1{n(2n-1)}\\
&\to c
\qquad\text{for some real }c\text{ since the sum converges}\\
s(2m+2)
&=s(2m+1)+ \frac{(-1)^{2m+2}(2m+2+1)}{2m+2-1}\\
&=s(2m+1)+ \frac{(2m+3)}{2m+1}\\
&=s(2m+1)+ 1+\frac{2}{2m+1}\\
&\to c+1\\
\end{array}
$
Therefore the even terms and odd sums
approach different limits,
so the series itself
does not have a limit.
Note:
this also works for
$\sum_{n=2}^{m} \frac{(-1)^n(n+a)}{n+b}
$
with $a \ne b$.
