Prove convergence / divergence of $\sum_{n=2}^\infty(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\sin\left(\frac {1}{\sqrt n}\right)$ How can I prove or disprove the following series converges?
$$\sum_{n=2}^\infty(-1)^n\cfrac {\sqrt n}{(-1)^n+\sqrt n}\sin\left(\frac {1}{\sqrt n}\right)$$
I tried several things, none of which worked. 
I wanted to use Abel's test or Dirichlet's test. I know that $\sin(\frac {1}{\sqrt n})$ is monotonically decreasing to $0$, but I wasn't able to show that $\Sigma_{n=2}^\infty(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}$ is convergent, as it does not converge absolutely since $\frac {\sqrt n}{(-1)^n+\sqrt n}$ converges to 1. Neither was I able to show that the partial sum sequence of $\frac {\sqrt n}{(-1)^n+\sqrt n}$ is bounded. I'm at a loss. Would love any help.
Note - This exact question was discussed here a few years ago, but was not answered then and the hint provided in the responses was not useful.
 A: 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?
A: 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 $-\infty$.
A: From ${1\over1+u}=1-u+{u^2\over1+u}$ we have
$${\sqrt n\over(-1)^n+\sqrt n}={1\over1+{(-1)^n\over\sqrt n}}=1-{(-1)^n\over\sqrt n}+{{1\over n}\over1+{(-1)^n\over\sqrt n}}=1-{(-1)^n\over\sqrt n}+{1\over n+(-1)^n\sqrt n}$$
Now, on multiplying by $(-1)^n\sin\left(1\over\sqrt n\right)$, we see that
$$\sum_{n=2}^\infty(-1)^n\sin\left(1\over\sqrt n\right)$$
converges conditionally while
$$\sum_{n=2}^\infty{(-1)^n\over n+(-1)^n\sqrt n}\sin\left(1\over\sqrt n\right)$$
converges absolutely (since $\sin(1/\sqrt n)/(n+(-1)^n\sqrt n)\approx1/n^{3/2}$ for large $n$), but
$$\sum_{n=2}^\infty{1\over\sqrt n}\sin\left(1\over\sqrt n\right)\approx\sum_{n=2}^\infty{1\over n}$$
diverges. So the overall sum diverges.
