Does the series of $\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}$ converge?

Question

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.

Answer 1

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

Answer 2

Hint:

What is $$\lim\limits_{n\to\infty}\frac{n+1}{n-1}? $$

Answer 3

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 $a \ne b$.