Proving that an alternating sum of arctangents is bounded

Question

I am struggling with the following question - how do I prove that a series like the following is bounded?

$$\sum_{n=1}^{ \infty} (-1)^n\arctan\left(\frac{3n^2}{2n+1}\right) $$

I just started learning about sums, so I'm trying to understand techniques to prove things like this. Thanks in advance :)

Answer 1

Hint. One may write $$ (-1)^n\arctan\left(\frac{3n^2}{2n+1}\right)=(-1)^n\frac \pi2 - (-1)^n\arctan\left(\frac{2}{3n}+\frac{1}{3n^2}\right) $$ then $$ \left|\sum_{n=1}^N(-1)^n\frac \pi2\right|\le \frac \pi2 \cdot \left|\sum_{n=1}^N(-1)^n\right|\le \frac \pi2 \cdot 2=\pi $$ and, by applying the Dirichlet test, $$ \left|\sum_{n=1}^\infty (-1)^n\arctan\left(\frac{2}{3n}+\frac{1}{3n^2}\right)\right|=C<\infty $$ giving $$ \left|\sum_{n=1}^\infty(-1)^n\arctan\left(\frac{3n^2}{2n+1}\right)\right|\le \pi+C. $$

Answer 2

As $n\to\infty $ we have $arctan(\frac{3n^2}{2n+1})\to \frac{\pi}{2}$ so this summation gets $\sum_{n \\ sufficiently \ \ \ large} (-1)^n\frac{\pi}{2}$ which is bounded but never converges.