Is $\sum_{n=1}^{\infty}\arctan{\frac{1}{n}}$ finite? As I have seen Showing that $ \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}$

Is $$\sum_{n=1}^{\infty}\arctan{\frac{1}{n}}&lt;\infty?$$
What is $$\lim_{h\to 0}\frac{\tan^{-1}(h)}h?$$
With Stolz-Cesaro and L'hopital rule:

$$L=\lim_{n\to \infty}\sum_{k=1}^{n}arc\tan\frac{1}{k}=\lim_{n\to \infty}\frac{\sum_{k=1}^nn \: arc\tan\frac{1}{k}}{n}=\lim_{n\to \infty}\frac{narc\tan\frac{1}{n}}{n-(n-1)}=\lim_{n\to \infty} narc\tan\frac{1}{n}=\lim_{x\to 0}\frac{arc\tan x}{x}=\lim_{x\to 0}\frac{1}{1+x^2}=1$$

Result: $L=1$