Find the value of the p for which the series converges

Question

$$\sum _{n=1}^{\infty \:}\frac{\arctan ^2\left(n\right)}{n^2\left(1+n^2\right)^{p}}$$

I would appreciate if anyone was able to help clear this up for me.

Answer 1

Looking at $\sum _{n=1}^{\infty \:}\frac{\arctan ^2\left(n\right)}{n^2}$, this series converges, which means that whenever $p\geq0$, your sum converges too by the comparison test.

So maybe it would be easier to look at $\sum _{n=1}^{\infty \:}\frac{(1+n^2)^k\arctan ^2\left(n\right)}{n^2}$, where $k=-p$.

When $k=0.5$, we have that $\sum _{n=1}^{\infty \:}\frac{(1+n^2)^{0.5}\arctan ^2\left(n\right)}{n^2}\geq \sum _{n=1}^{\infty \:}\frac{(n^2)^{0.5}\arctan ^2\left(n\right)}{n^2}=\sum _{n=1}^{\infty \:}\frac{\arctan ^2\left(n\right)}{n} $,which diverges.

When $k<0.5$, we have that $\sum _{n=1}^{\infty \:}\frac{(1+n^2)^{k}\arctan ^2\left(n\right)}{n^2} \leq \sum _{n=1}^{\infty \:}\frac{(n^2+n^2)^{k}\arctan ^2\left(n\right)}{n^2}=2^k\sum _{n=1}^{\infty \:}\frac{\arctan ^2\left(n\right)}{n^{2-2k}}$, which converges whenever $k<0.5$.