Determine the convergence of the following series $\sum_{n=2}^{\infty} (-1)^n \frac{1}{\sqrt{n} + (-1)^n}$ $\sum_{n=2}^{\infty} (-1)^n \frac{1}{\sqrt{n} + (-1)^n}$
Now i know that this is alternating series which means that i should determine the  absolute convergence $|a_n|=\frac{1}{\sqrt{n} + (-1)^n}$ but i don't know how to do it, d'Alembert's test isn't working Cauchy tests can't help here here either so i am out of ideas now. 
 A: $$\begin{align}
\frac{(-1)^n}{\sqrt{n}+(-1)^n}&=\frac{(-1)^n}{\sqrt{n}}+\Bigl(\frac{(-1)^n}{\sqrt{n}+(-1)^n}-\frac{(-1)^n}{\sqrt{n}}\Bigr)\\
&=\frac{(-1)^n}{\sqrt{n}}-\frac{1}{\sqrt{n}\,\bigl(\sqrt{n}+(-1)^n\bigr)}.
\end{align}$$
Now:


*

*$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt{n}}$ converges by Leibniz's criterion.

*$\displaystyle\sum_{n=2}^\infty\frac{1}{\sqrt{n}(\sqrt{n}+(-1)^n)}$ diverges.


The original series diverges.
Another way of seeing it is to observe that the sum of two consecutive terms behaves like $1/n$, so that the even partial sums do not converge.
A: Note that
$$\sum_{n=2}^m\frac{(-1)^n}{\sqrt{n} + (-1)^n}= \sum_{n=2}^m\frac{(-1)^n(\sqrt{n} - (-1)^n)}{(\sqrt{n} + (-1)^n)(\sqrt{n} - (-1)^n)} \\ =\sum_{n=2}^m\frac{(-1)^n\sqrt{n}-1}{n -1}\\ = \sum_{n=2}^m\frac{(-1)^n\sqrt{n}}{n -1}- \sum_{n=2}^m\frac{1}{n -1}. $$
The first sum converges by Dirichlet and the second sum diverges.
Hence the series diverges.
A: Since $$\frac 1{\sqrt n + (-1)^n} \sim \frac 1{\sqrt n}$$ and the latter series does not converge, neither does your $|a_n|$
a problem with your approach though is that since the absolute series does not converge, you can't conclude anything about the series with term $a_n$
A: Compare the first two terms, $ \frac{1}{\sqrt{2} +1}, \frac{1}{\sqrt{3}-1}$: the sequence is not monotone decreasing, so the alternating series test does not apply
