Convergent series

Study the convergence of the series $$\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n}$$



Solution. Rewrite in the form $$\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{2n}}{\sqrt{2n}+(-1)^{2n}}+\frac{(-1)^{2n+1}}{\sqrt{2n+1}+(-1)^{2n+1}}\right)$$ (essentially grouping the summand with even power of $-1$ with the consequent one)

Now the general term could be rewritten in the form $$\frac{-2 - \sqrt{2n} + \sqrt{
 1 + 2 n}}{(1 + \sqrt{2n}) (-1 +\sqrt{1 + 2 n})}$$

These are all negative. The numerator tends to $-2$ (since $- \sqrt{2n} + \sqrt{
 1 + 2 n}\longrightarrow 0$). The denominator is of order $2n$, thus $$\frac{-2 - \sqrt{2n} + \sqrt{
 1 + 2 n}}{(1 + \sqrt{2n}) (-1 +\sqrt{1 + 2 n})}\sim \frac{-1}{n}$$
Thus the series is convergent if and only if the series $$\sum_{n=1}^{\infty}\frac{-1}{n}$$ is convergent, but the latter diverges, hence the original series diverges.