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.
$$\begin{align}
\frac{(-1)^n}{\sqrt{n}+(-1)^n}&amp;=\frac{(-1)^n}{\sqrt{n}}+\Bigl(\frac{(-1)^n}{\sqrt{n}+(-1)^n}-\frac{(-1)^n}{\sqrt{n}}\Bigr)\\
&amp;=\frac{(-1)^n}{\sqrt{n}}-\frac{1}{\sqrt{n}\,\bigl(\sqrt{n}+(-1)^n\bigr)}.
\end{align}$$
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.
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