Test the series for conditional convergence: $ \sum_{n=1}^{\infty}a_n,\ \ a_n=\arctan\frac{\sin n}{\sqrt{n+1}} $ Test the series for conditional convergence:
$$
\sum_{n=1}^{\infty}a_n,\ \ a_n=\arctan\frac{\sin n}{\sqrt{n+1}}
$$
I tried to apply the fact that $|\arctan(t)|\le|t|\ \ \forall |t|<1$ and then prove that the initial series converges absolutely (and therefore conditionally).
However, this idea didn't work:
$$
|a_n|<\left|\frac{\sin n}{\sqrt{n+1}}\right|=\frac{|\sin n|}{\sqrt{n+1}}=b_n
$$
And the series $\sum_{n=1}^\infty b_n$ is divergent.
Every other method I know of doesn't help here, because $a_n$ can take both positive and negative values. Dirichlet test doesn't work either, since I have one unsplittable function here, which is $\arctan$.
So, what should I do?
 A: $$\sum_{n=1}^\infty \arctan \frac{\sin n}{\sqrt{n+1}} = \sum_{n=1}^\infty  \frac{\sin n}{\sqrt{n+1}} + \sum_{n=1}^\infty \left(\arctan \frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}}\right)$$
Dirichlet test for the first, and the second converges absolutely.
A: Hint (almost solution):
Statement: $\sum_{n=1}^{\infty}$ converges conditionally.
It's know that $|\arctan(x) - x| \le 100 |x|^3$ for $-\frac{1}{100} \le x \le  \frac{1}{100}$.
A series $100\sum_{n=1}^{\infty} (\frac{\sin n}{\sqrt{n+1}})^3$ converges absolutely, because $|(\frac{\sin n}{\sqrt{n+1}})^3 | \le \frac{1}{n^{\frac32}}$. Hence $\sum_{n=1}^{\infty} (\arctan\frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}})$ converges absolutely.
As
$$\sum_{n=1}^{\infty} \arctan\frac{\sin n}{\sqrt{n+1}} = \sum_{n=1}^{\infty} (\arctan\frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}}) + \sum_{n=1}^{\infty} \frac{\sin n}{\sqrt{n+1}},$$
it's sufficient to show that $\sum_{n=1}^{\infty} \frac{\sin n}{\sqrt{n+1}}$ converges conditionally. We may get convergence from Dirichlet's test for convergence.
It's sufficient to show that $\sum_{n=1}^{\infty} \big|\frac{\sin n}{\sqrt{n+1}} \bigr| = +\infty$.
