Does the series $\sum_{k=1}^{\infty}(\arctan(\frac{1}{k}))^{2}$ converge or diverge?

Solution 1:

Yes, you can compare $\left(\arctan^2\left(\frac1k\right)\right)_{k\in\Bbb N}$ with $\left(\frac1{k^2}\right)_{k\in\Bbb N}$. And you can do it using the fact that\begin{align}\lim_{k\to\infty}\frac{\arctan^2\left(\frac1k\right)}{\frac1{k^2}}&=\lim_{k\to\infty}\left(\frac{\arctan\left(\frac1k\right)}{\frac1k}\right)^2\\&=\left(\lim_{k\to\infty}\frac{\arctan\left(\frac1k\right)}{\frac1k}\right)^2\\&=1.\end{align}So, by the comparison test, and since $\sum_{k=1}^\infty\frac1{k^2}$ converges and both series are series of numbers greater than $0$, the series$$\sum_{k=1}^\infty\arctan^2\left(\frac1k\right)$$converges.