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?
Best Answer
$$\sum_{n=1}^\infty \arctan \frac{\sin n}{\sqrt{n+1}} = \sum_{n=1}^\infty \frac{\sin n}{\sqrt{n+1}} + \sum_{n=1}^\infty \left(\arctan \frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}}\right)$$ Dirichlet test for the first, and the second converges absolutely.