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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?

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    $\begingroup$ Try using the power series for $\arctan(x)$. You can split the resulting series into two parts. $\endgroup$ – Clayton Dec 24 '20 at 18:05
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$$\sum_{n=1}^\infty \arctan \frac{\sin n}{\sqrt{n+1}} = \sum_{n=1}^\infty \frac{\sin n}{\sqrt{n+1}} + \sum_{n=1}^\infty \left(\arctan \frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}}\right)$$ Dirichlet test for the first, and the second converges absolutely.

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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 $\sum_{n=1}^{\infty} \big|\frac{\sin n}{\sqrt{n+1}} \bigr| = +\infty$.

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  • $\begingroup$ Great supplement to Robert's answer! Thank you! $\endgroup$ – Bonrey Dec 24 '20 at 18:21
  • $\begingroup$ Could you explain why $|\arctan(x)-x|<Cx^3,\ \ -\frac{1}{C}<x<\frac{1}{C}$ ? $\endgroup$ – Bonrey Dec 24 '20 at 18:22
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    $\begingroup$ @Bonrey, it was only for "big" $C$ such as $C=100$. We know that $\frac{arctan x - x}{x^3} \to \frac{1}{3}$, $x \to 0$, and using standard methods we may get that $| \frac{arctan x - x}{x^3} - \frac{1}{3}| \le 50$ for $x \in [- \frac{1}{100}, \frac{1}{100}]$ and hence $|arctan x - x| < 100 |x|^3$. Number $C = 100$ is not essential - you may use any other "big" $const$. $\endgroup$ – Botnakov N. Dec 24 '20 at 18:31

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