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.
A: This is a rearrangement of   the Alternating series, but a rather simple one. What it amounts to is grouping terms of the Alternating series and rearranging the terms within each group.
Naively:
$$\eqalign{
1-{1\over 2}+ {1\over 3}-{1\over4}+\cdots
&=
1-{1\over 2}+ {1\over 3}-{1\over4} +\cdots\cr
&=\Bigl( 1-{1\over 2}  \Bigr) +\Bigl( {1\over 3}-{1\over4}\Bigr) +\cdots\cr
&=\Bigl(  -{1\over 2} +1 \Bigr) +\Bigl( -{1\over 4}+{1\over3}\Bigr) +\cdots\cr
}
$$ 
Is this justified? Yes it is:
Let $S_n$ denote the $n$'th Partial sum of the Alternating series and $T_n$ denote the $n$'th partial sum of the OP's rearranged series. Then for $n>1$, we have $|S_n-T_n| \le{1\over n }-{1\over n+1}={1\over n(n+1)}$. Thus for any $n$
$$
S_n-{1\over n( n+1)}\le T_n \le S_n+{1\over n(n+1)}.
$$
From this and the fact that $S_n$ converges, it follows that $T_n$ converges (and to the same limit as $S_n$).
A: Hint: let $a_n=\frac{(-1)^n}{n+(-1)^{n+1}}$ and let $(b_1,b_2,b_3,b_4,b_5,b_6,\ldots)=(a_2,a_1,a_4,a_3,a_6,a_5,\ldots)$, then $\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$, which is convergent by the alternating series test. So, ...
