# Is $\sum_{n=1}^{\infty}\arctan{\frac{1}{n}}$ finite?

Is $$\sum_{n=1}^{\infty}\arctan{\frac{1}{n}}<\infty?$$

Look at a graph of $\arctan$ near $x=0$. It's asymptotically equivalent to $y=x$. So as $n\to\infty$, $\frac{1}{n}\to0$, and $\arctan\left(\frac1n\right)\approx\frac{1}{n}$. This series is comparable to the harmonic series.

What is $$\lim_{h\to 0}\frac{\tan^{-1}(h)}h?$$

• I don't get the point... However, (arctan x)'=1/(1+x^2)...... @Pedro Tamaroff Commented Dec 5, 2013 at 16:12
• @user39843 Use the "comparison test".
– Pedro
Commented Dec 5, 2013 at 16:15
• Oh... I see! Thanks! Commented Dec 5, 2013 at 16:16

I am assuming you know some convegence test.We have to find whether this series converge or not.since we have $$\lim_{n\to{\infty}}\frac{tan^{-1}\frac{1}{n}}{\frac{1}{n}}=1\neq0$$.Hence By limit comparisson test $\sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum_{n=1}^{\infty} tan^{-1}\frac{1}{n}$ have same behaviour.But $\sum_{n=1}^{\infty}\frac{1}{n}$ is diverging.So $\sum_{n=1}^{\infty} tan^{-1}\frac{1}{n}$ also diverges.So $$\sum_{n=1}^{\infty} tan^{-1}\frac{1}{n}$$ is not finite.

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: $L=1$