# 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?

• Try using the power series for $\arctan(x)$. You can split the resulting series into two parts. – Clayton Dec 24 '20 at 18:05

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

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

• Great supplement to Robert's answer! Thank you! – Bonrey Dec 24 '20 at 18:21
• Could you explain why $|\arctan(x)-x|<Cx^3,\ \ -\frac{1}{C}<x<\frac{1}{C}$ ? – Bonrey Dec 24 '20 at 18:22
• @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" $const$. – Botnakov N. Dec 24 '20 at 18:31