# 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.

• How about the $n$th term test for divergence. Commented Mar 5, 2015 at 20:22

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

$$L-1 < \sum_{n=2}^{2N+1}(-1)^n \frac{n+1}{n-1} < L-\frac{1}{3}$$

This is the idea of the proof of why the general term must converge to 0

• Sometime it's good to remember why the general term of a serie must converge to 0 ;) Commented Mar 5, 2015 at 20:48

Hint:

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

• I think a more accurate hint would be "What is $\lim_{n\to\infty} (-1)^n\frac{n+1}{n-1}$?" That directly invokes the fact that in a convergent series, the summand must tend to $0$. Commented Mar 5, 2015 at 20:23
• @GregMartin I considered that, but I chose this since this limit actually exists. Also, you gave away the hint! ;-)
– Eff
Commented Mar 5, 2015 at 20:25
• I know "limit((n+1)/(n-1),x = infinity) = 1" Commented Mar 5, 2015 at 20:30
• @user221318 Correct. Do you know what is required for an infinite series? Particularly, do you know what is required of the summand for the series to converge?
– Eff
Commented Mar 5, 2015 at 20:31
• that they decrease as n-> infinity Commented Mar 5, 2015 at 20:33

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 $$a \ne b$$.