check if series $\sum^\infty_{n=1} \frac{(-1)^n}{\sqrt{n}+{(-1)^n}}$ converges There's the series:
$$\sum^\infty_{n=1} \frac{(-1)^n}{\sqrt{n}{+(-1)^n}}$$
I have no clue which test should I use to check it properly
Root test gives nothing right as well as Ratio test. I suspect there's a trap
 A: The series $\sum_{n=2}^N\frac{(-1)^n}{\sqrt{n}+(-1)^n}$ diverges.
To show this we write the partial sums of the series as
$$\begin{align}
\sum_{n=2}^N\frac{(-1)^n}{\sqrt{n}+(-1)^n}&=\color{blue}{\underbrace{\sum_{n=2}^N\frac{(-1)^n}{\sqrt{n}}}_{\text{Converges as}\,\,N\to \infty}}-\color{red}{\underbrace{\sum_{n=2}^N\left(\frac{1}{\sqrt{n}(\sqrt{n}+(-1)^n)}\right)}_{\text{Diverges as}\,\,N\to \infty}}\tag 1
\end{align}$$
Leibniz's test guarantees that the limit first term converges as $N\to \infty$.  
However, for the second term on the right, we have
$$\sum_{n=2}^N \frac{1}{\sqrt{n}(\sqrt{n}+(-1)^n)}\ge \frac12 \sum_{n=2}^N\frac1{n}$$
which shows that the second term diverges by comparison to the harmonic series.

Inasmuch as the partial sums of the series of interest are comprised of the partial sums of a convergent series and the partial sums of a divergent series, the series of interest diverges.  

A: For the new version
$$
S=\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}+(-1)^n}
$$
The alternating series does not apply since the terms do not decrease in absolute value.  However, group the terms in twos
$$
\sum_{k=1}^\infty \left(\frac{(-1)^{2k-1}}{\sqrt{2k-1}+(-1)^{2k-1}} +
\frac{(-1)^{2k}}{\sqrt{2k}+(-1)^{2k}}\right)
$$
but this term is
$$
\left(\frac{(-1)}{\sqrt{2k-1}+(-1)} +
\frac{(+1)}{\sqrt{2k}+(+1)}\right)
=\frac{\sqrt{2k-1}-1-\sqrt{2k}-1}{\big(\sqrt{2k-1}-1\big)\big(\sqrt{2k}+1\big)} \\ <
\frac{-1}{\big(\sqrt{2k-1}-1\big)\big(\sqrt{2k}+1\big)} < 0
$$
These terms are all negative, so you can show divergence
by comparison to $\sum (-1/k)$.
