Q:

About $\sum_{n=1}^{\infty}\frac{(-1)^n}{n+(-1)^{n+1}}$

Determine if the following series is converges or diverges
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n+(-1)^{n+1}}$$  
I suspecting that we need rewrite $ \frac{(-1)^n}{n+(-1)^{n+1}} $ somehow, but how? 
Thanks! 

A:

It often helps to write out a few terms to try to see what’s going on:
$$-\frac12+\frac11-\frac14+\frac13-\frac16+\frac15-\frac18+\frac17-+\ldots$$
If you calculate even-numbered partial sums, you’re going to get sums like
$$s_8=\left(-\frac12+\frac11\right)+\left(-\frac14+\frac13\right)+\left(-\frac16+\frac15\right)+\left(-\frac18+\frac17\right)\;,$$
for instance.  With a bit of thought you can work out that
$$\begin{align*}
s_{2n}&=\left(-\frac12+\frac11\right)+\left(-\frac14+\frac13\right)+\ldots+\left(-\frac1{2n}+\frac1{2n-1}\right)\\\\
&=\sum_{k=1}^n\left(-\frac1{2k}+\frac1{2k-1}\right)\\\\
&=\sum_{k=1}^n\frac1{2k(2k-1)}\;,
\end{align*}$$
and
$$s_{2n+1}=s_{2n}-\frac1{2(n+1)}=\sum_{k=1}^n\frac1{2k(2k-1)}-\frac1{2(n+1)}\;.$$
The original series converges if the sequence of these partial sums converges.  Concentrate on the even-numbered ones, and worry about the odd-numbered ones afterwards.  Does
$$\sum_{k=1}^\infty\frac1{2k(2k-1)}$$
converge?  If it doesn’t, the sequence of even-numbered partial sums of your original series doesn’t converge, so the series cannot converge.  If it does, the sequence of even-numbered partial sums of your original series does converge, and you have only to check that the odd-numbered partial sums don’t misbehave.