Is $\sum_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}$ absolute convergent,conditional convergent or divergent? Is $\sum_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}$ absolute convergent,conditional convergent or divergent ?
$\sum_{n=2}^{\infty} |\frac{(-1)^n }{n^2+(-1)^n}| =\frac{1}{|n^2+(-1)^n|}$
now $|n^2+(-1)^n| > n^2 - 2 $ 
=>$\frac{1}{|n^2+(-1)^n|}<1/(n^2 - 2)$
$1/(n^2 - 2)$ is convergent by Limit comparison test with $1/n^2$
therefore,$\frac{1}{|n^2+(-1)^n|}$ is convergent by comparison test
Therefore, case of absolute convergence?
is this correct ?

$\sum\limits_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}=\sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2+1}-\sum\limits_{n=1}^{\infty} \frac{1}{(2n+1)^2-1}$ 
with
$\sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2+1}\leq \frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{24}$
and 
$\sum\limits_{n=1}^{\infty} \frac{1}{(2n+1)^2-1}=\frac{1}{4}\sum\limits_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1})=\frac{1}{4}$
Therefore 
$|\sum\limits_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}|\leq \sum\limits_{n=2}^{\infty} \frac{1}{n^2+(-1)^n}\leq \frac{\pi^2}{24}+\frac{1}{4}$
=> absolute convergence
