determine whether or not infinite series $ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n} $ converge I want to determine if the series $ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n} $ converge/diverge. the sequence in the denominator is not monotinic, so I cant use Dirichlet's or Abel's tests. My intuition is that this series converge, becuase its looks close to $ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n} $ but im not sure how to prove. Any ideas will help, thanks.
 A: Let 
$$s_n=\sum_{k=2}^n\frac{(-1)^k}{(-1)^k+k}=\frac13-\frac12+\frac15-\frac14+\ldots+\frac{(-1)^n}{(-1)^n+n}$$ 
and 
$$s_n'=\sum_{k=2}^n\frac{(-1)^{k+1}}k=-\frac12+\frac13-\frac14+\frac15+\ldots+\frac{(-1)^{n+1}}n\;.$$
Show that $s_{2n+1}=s_{2n+1}'$ and $s_{2n}=s_{2n+1}'+\frac1{2n}$ for $n\ge 1$. Use this or the fact that $s_{2n}=s_{2n}'+\frac1{2n}+\frac1{2n+1}$ to show that $\lim_\limits{n\to\infty}|s_n-s_n'|=0$, and therefore $\lim_\limits{n\to\infty}s_n=\lim_\limits{n\to\infty}s_n'$.
A: Let
$s(m)
=\sum_{n=2}^{m}\dfrac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n}
$.
The terms go to zero,
so it enough to show that
$s(2m+1)$
converges.
$\begin{array}\\
s(2m+1)
&=\sum_{n=2}^{2m+1}\dfrac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n}\\
&=\sum_{n=1}^{m}(\dfrac{\left(-1\right)^{2n}}{\left(-1\right)^{2n}+2n}+\dfrac{\left(-1\right)^{2n+1}}{\left(-1\right)^{2n+1}+2n+1})\\
&=\sum_{n=1}^{m}(\dfrac{1}{1+2n}+\dfrac{-1}{-1+2n+1})\\
&=\sum_{n=1}^{m}(\dfrac{1}{1+2n}-\dfrac{1}{2n})\\
&=\sum_{n=1}^{m}\dfrac{-1}{2n(2n+1)}\\
\end{array}
$
and this sum converges
by comparison with
$\sum \dfrac1{4n^2}
$.
To get an explicit bound,
$\begin{array}\\
-s(2m+1)
&=\sum_{n=1}^{m}\dfrac{1}{2n(2n+1)}\\
&=\dfrac16+\sum_{n=2}^{m}\dfrac{1}{2n(2n+1)}\\
&<\dfrac16+\sum_{n=2}^{m}\dfrac{1}{2n(2n-2)}\\
&=\dfrac16+\dfrac14\sum_{n=2}^{m}\dfrac{1}{n(n-1)}\\
&=\dfrac16+\dfrac14\sum_{n=2}^{m}(\dfrac{1}{n-1}-\dfrac{1}{n})\\
&=\dfrac16+\dfrac14(1-\dfrac1{m})\\
&< \dfrac{7}{12}\\
\text{and}\\
-s(2m+1)
&=\sum_{n=1}^{m}\dfrac{1}{2n(2n+1)}\\
&=\dfrac16+\sum_{n=2}^{m}\dfrac{1}{2n(2n+1)}\\
&>\dfrac16+\sum_{n=2}^{m}\dfrac{1}{2n(2n+2)}\\
&=\dfrac16+\dfrac14\sum_{n=2}^{m}\dfrac{1}{n(n+1)}\\
&=\dfrac16+\dfrac14\sum_{n=2}^{m}(\dfrac1{n}-\dfrac1{n+1})\\
&=\dfrac16+\dfrac14(\frac12-\dfrac1{m+1})\\
&=\dfrac16+\dfrac18-\dfrac1{4(m+1)})\\
&=\dfrac{7}{24}-\dfrac1{4(m+1)}\\
\end{array}
$
A: The series $\sum_{n=2}^{\infty}\frac{(-1)^n}{n}$ converges by the alternating series test. $$\text{Your given series }-\sum_{n=2}^{\infty}\frac{(-1)^n}{n}=-\sum_{n=2}^{\infty}\frac{1}{n((-1)^n+n)}$$The series $$\sum_{n=2}^{\infty}\frac{1}{n((-1)^n+n)}$$ converges by the limit comparison test with the convergent $p-$series 
$$\sum_{n=2}^{\infty}\frac{1}{n^2}.$$ Thus your given series is the difference of two convergent series and hence your given series also converges.
A: Let $ n $ be a positive integer.
\begin{aligned}\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n}}&=\frac{\left(-1\right)^{n}}{n}\left(\frac{n}{n+\left(-1\right)^{n}}\right)\\ &=\frac{\left(-1\right)^{n}}{n}\left(1-\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n}}\right)\\ &=\frac{\left(-1\right)^{n}}{n}+v_{n}\end{aligned}
Where $ v_{n}=-\frac{1}{n^{2}+n\left(-1\right)^{n}}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) \cdot $
Since $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{n}} $ converges, and $ \sum\limits_{n\geq 1}{v_{n}} $ converges by comparaison, we get that $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n}}} $ converges.
A: It sometimes helps to write out the first few terms, in order to see what you're dealing with, and possibly spot a useful pattern. In this case we have
$$\begin{align}
\sum_{n=2}^\infty{(-1)^n\over(-1)^n+n}
&={1\over3}-{1\over2}+{1\over5}-{1\over4}+{1\over7}-{1\over6}+\cdots\\
&=-\left({1\over2}-{1\over3}+{1\over4}-{1\over5}+{1\over6}-{1\over7}+\cdots \right)
\end{align}$$
Now depending on your standard of rigor, that may already by enough to prove conditional convergence. If you need to be more finicky, then a careful examination of the expansion tells us
$$\sum_{n=2}^N{(-1)^n\over(-1)^n+n}=-\sum_{n=2}^N{(-1)^n\over n}+
\begin{cases}
0&\text{if $N$ is odd}\\
\displaystyle{1\over N}-{1\over N+1}&\text{if $N$ is even}
\end{cases}$$
Since $\sum(-1)^n/n$ is conditionally convergent by the familiar tests, and since ${1\over N}-{1\over N+1}\to0$ as $N\to\infty$, the given series converges (conditionally) as well.
