Does $\sum_{n=1}^{\infty}\arctan(\frac{1}{n^2})$ converge? I tried to prove that the series $\sum_{n=1}^{\infty}\arctan(\frac{1}{n^2})$ converges. I tried using $\lim_{n\to \infty}\frac{\arctan(\frac{1}{n^2})}{\frac{1}{n^2}}$ as we knew that  $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges but the limit went to $\infty$,
So I can't say anything about $\sum_{n=1}^{\infty}\arctan(\frac{1}{n^2})$. It is hard even to try to find  the integral of $\arctan(\frac{1}{n^2})$. Is there another way?
 A: Note that$$\lim_{n\to\infty}\frac{\arctan\left(\frac1{n^2}\right)}{\frac1{n^2}}=\arctan'(0)=1.$$Therefore, your approach actually works.
A: hint: $\arctan x < x, x \in (0,\pi/4)$
A: You're wrong: $\;\arctan\dfrac1{n^2}\sim_\infty\dfrac1{n^2}$, as can easily be seen with Taylor's expansion at order $1$. The latter is a convergent Riemann $p$ series.
A: Yes the approach is correct and since
$$\lim_{n\to \infty}\frac{\arctan(\frac{1}{n^2})}{\frac{1}{n^2}}=1$$
we can conclude that the given series converges by limit comparison test with $\sum_{n=1}^{\infty}\frac{1}{n^2}$.
A: The sum is not only convergent but it has a nice closed form
$$s_{c2} = \sum _{n=1}^{\infty } \arctan \left(\frac{1}{n^2}\right)\\=
\arctan \left(\frac{1-\cot \left(\frac{\pi }{\sqrt{2}}\right) \tanh \left(\frac{\pi }{\sqrt{2}}\right)}{1+\cot \left(\frac{\pi }{\sqrt{2}}\right) \tanh \left(\frac{\pi }{\sqrt{2}}\right)}\right)\\\simeq 1.42474...$$
It can be derived by the method described in https://math.stackexchange.com/a/2987779/198592.
