Prove convergence / divergence of $\sum_{n=2}^\infty(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\sin\left(\frac {1}{\sqrt n}\right)$

Hint. By Taylor series expansions, one has, as $x \to 0$, $$ \sin x=x+O(x^3), \qquad \frac1{1+x}=1-x+O(x^2), $$giving, as $n \to \infty$, $$ (-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}=\frac {(-1)^n}{1+\frac{(-1)^n}{\sqrt n}}=(-1)^n-\frac{1}{\sqrt n}+O\left(\frac {1}{n} \right), $$and$$ \sin\frac {1}{\sqrt n}=\frac {1}{\sqrt n}+O\left(\frac {1}{n^{3/2}} \right). $$ Then, as $n \to \infty$, one gets

$$ (-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\:\sin \frac {1}{\sqrt n}=\frac{(-1)^n}{\sqrt n}-\frac1n+O\left(\frac {1}{n^{3/2}} \right). $$

Can you take it from here?

Observe that $$ \frac{\sqrt{n}}{\sqrt{n} + (-1)^n} = \frac{\sqrt{n}}{\sqrt{n} + (-1)^n} \cdot \frac{\sqrt{n} - (-1)^n}{\sqrt{n} - (-1)^n} = \frac{n - (-1)^n \sqrt{n}}{n-1}, $$ hence the general term $a_n$ of your series can be written as $$ a_n = b_n + c_n, \qquad b_n := (-1)^n \frac{n}{n-1} \sin \frac{1}{\sqrt{n}}, \quad c_n := - \frac{\sqrt{n}}{n-1} \sin \frac{1}{\sqrt{n}}\,. $$ Now, $\sum_n b_n$ is convergent by Leibnitz's criterion for alternating series, whereas $\sum c_n$ diverges to $-\infty$ (since $c_n \sim - 1/n$). So $\sum a_n$ diverges to $-\infty$.