Prove the convergence of $\sum \frac{(-1)^n}{n+(-1)^n}$ How would one go around proving the convergence of: $ \sum \frac{(-1)^n}{n+(-1)^n} $. I'm fairly certain, that this series converges, but it doesn't do so absolutely $\left( \frac{1}{n+(-1)^n} \approx \frac{1}{n}\right)$. Leibniz criterion of convergence can't be used, because  $\left( \frac{1}{n+(-1)^n} \right) $ is not nonincreasing.
Note: I've noticed, that by shuffling terms of the series one would get $\sum (-1)^n a_n$, where $a_n$ is a nonincreasing sequence, but shuffling is not allowed since the series doesn't converge absolutely. 
 A: Note that 
$$\frac{(-1)^n}{n+(-1)^n}=\begin{cases}\frac{1}{2k+1}&\text{if $n=2k$,}\\
-\frac{1}{2k}&\text{if $n=2k+1$.}\end{cases}$$
Hence
$$\sum_{n=2}^{N}\frac{(-1)^n}{n+(-1)^n}=
\begin{cases}\sum_{k=2}^{N}\frac{(-1)^{k+1}}{k}+\frac{1}{N}+\frac{1}{N+1}&\text{if $N$ is even,}\\
\sum_{k=2}^{N}\frac{(-1)^{k+1}}{k}&\text{if $N$ is odd.}\end{cases}$$
Therefore the given series is convergent and it is convergent to the same sum of $\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k}$ (convergent by Leibniz):
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n+(-1)^n}=\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k}=-1+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=-1+\ln(2).$$
A: \begin{align}\sum_{k=2}^n\frac{(-1)^n}{n+(-1)^n}&=\frac13-\frac12+\frac15-\frac14+\frac17-\frac16+\cdots\\&=-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\cdots\end{align}and this series converges, by Leibniz criterion. This kind of shuffling is allowed. By “this kind” what I mean is the replacement of a series $\sum_{n=1}^\infty a_n$ by $\sum_{n=1}^\infty a_{\sigma(n)}$ where $\sigma\colon\mathbb{N}\longrightarrow\mathbb N$ is a bijection such that the set $\{n-\sigma(n)\,|\,n\in\mathbb{N}\}$ is bounded.
A: Notice that
$$ \frac{(-1)^n}{n+(-1)^n}
= \frac{(-1)^n}{n} - \frac{1}{n(n+(-1)^n)}.$$
Now


*

*$\sum \frac{(-1)^n}{n}$ converges by alternating series test, and 

*$ \sum \frac{1}{n(n+(-1)^n)}$ converges by comparison with $\sum \frac{1}{n^2} < \infty$.


Therefore their difference also converges as well. One may also identify this trick as the following asymptotic expansion
$$ \frac{(-1)^n}{n+(-1)^n}
= \frac{(-1)^n}{n\left(1 + \mathcal{O}(n^{-1}) \right)}
= \frac{(-1)^n}{n} + \mathcal{O}(n^{-2}) $$
which is summable.
A: $$\frac{(-1)^n}{n+(-1)^n}<\frac{(-1)^n}{2 n}, \text{ for any }n\in\mathbb{N}\land n\geq 2$$
As the second series converges for Leibniz criterion, so does the first
Hope this helps
A: $a_n: \dfrac{(-1)^n}{n+(-1)^n}= $
$\dfrac{(-1)^n}{n+(-1)^n} \dfrac{n-(-1)^n}{n-(-1)^n}=$
$\dfrac{(-1)^nn -1}{ n^2 -1} ;$
$b_n:= \dfrac{n}{n^2-1}=$
$ 1/2(\dfrac{1}{n+1} +\dfrac{1}{n-1});$
$c_n := -\dfrac{1}{n^2-1}.$
$a_n = (-1)^n b_n + c_n.$
Use Leibniz criterion for alternating series 
$\sum (-1)^nb_n$;
Use comparison test $(1/n^2)$ for 
$\sum c_n .$
Hence $ \sum a_n $ convergent.
