Convergence test for series $\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}\sqrt{n+(-1)^n}}$ What would be the approach to resolve whether this series converges (absolutely or conditionally) or diverges? 
 A: Denote $\sum a_n$ the given series. We have
$$\vert a_n\vert\ge\frac{1}{n+1}$$
so $\sum a_n$ isn't absolutely convergent. Moreover
$$a_n=\frac{(-1)^n}n\left(1+\frac{(-1)^n}{n}\right)^{-1/2}=\frac{(-1)^n}{n}+O\left(\frac1{n^2}\right)$$
so $\sum a_n$ is convergent as sum of two convergent series.
A: Hint:
$$\frac{1}{\sqrt {n+1} \sqrt { n+1 +(-1)^{n+1}} } \le \frac{1}{\sqrt {n+1}\sqrt n} \le \frac{1}{\sqrt {n}\sqrt { n +(-1)^n }}.$$
A: Let $S_n$ denote the sequence of partial sums and $a_n$ denote the general term of the series. Then, $S_n \to 0$: notice that $a_{2n} = - a_{2n + 1}$ for all $n \ge 1$. So,
$$S_{2n} = \sum_{k=2}^{2n} a_k = a_2 - a_2 + a_4 - a_4 + ... + a_{2n- 2} - a_{2n - 2} + a_{2n} = a_{2n} \to 0$$
$$S_{2n + 1} = \ldots = 0 \to 0$$
(if a sequence $(x_n)$ is such that the subsequences $(x_{2n})$ and $(x_{2n + 1})$ converge to the same number, then $(x_n)$ converges to that number)
A: The Dirichlet's test states that: if $a_n,b_2$ sequences of real numbers satisfy that: 
1- $a_n \geq a_{n+1}$
2- $\lim_{n \rightarrow \infty}a_n = 0$
3- $\left|\sum^{N}_{n=1}b_n\right|\leq M$ for every positive integer $N$
where $M$ is some constant, then the series
$\sum^{\infty}_{n=1}a_n b_n$ converges.
Take $a_n=\frac{1}{n}$ and $b_n=
\frac{{\left( { - 1} \right)^n }}{{\sqrt {1 + {\textstyle{{\left( { - 1} \right)^n } \over n}}} }}$.
Clearly, the conditions on $a_n$ is satisfied and there exists $M>0$
s.t.
$$
\left| {\sum\limits_{n = 2}^N {b_n } } \right| = \left| {\sum\limits_{n = 2}^N {\frac{{\left( { - 1} \right)^n }}{{\sqrt {1 + {\textstyle{{\left( { - 1} \right)^n } \over n}}} }}} } \right| \le \sum\limits_{n = 2}^N {\frac{{\left| {\left( { - 1} \right)^n } \right|}}{{\left| {\sqrt {1 + {\textstyle{{\left( { - 1} \right)^n } \over n}}} } \right|}}}  = \sum\limits_{n = 2}^N {\frac{1}{{\sqrt {1 + {\textstyle{{\left( { - 1} \right)^n } \over n}}} }}}  \le M
$$
(choose e.g. $M\ge 100$ or anything $\,\,\ge 2$)
