Determining whether the series: $\sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right)$ converges

I was tasked with determining whether the following series:

$$\sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right)$$

converges.

I tried employing the integral test which failed and produced incalculable integrals. Other methods didn't prove effective also. I was suggested that the Maclauren series might be of use here, but I'm not sure how to employ it.

Or by limit comparison test with $\sum\frac1n$ since by standard limit for $x\to 0\implies\frac{\tan x}{x}\to 1$ and then

$$\frac{\tan\left(\frac{1}{n}\right)}{\frac1n}\to1$$

• That's an efficient way to do it. Commented Mar 29, 2018 at 13:21
• @Bak1139 Yes indeed limit comparison test avoid to use inequalities which sometimes are difficult to handle.
– user
Commented Mar 29, 2018 at 13:23

We can solve this with the inequality $$\tan(x)>x$$ for $$0 as follows $$\sum_{n=1}^\infty \tan\left(\frac{1}{n}\right)\ge\sum_{n=1}^\infty\frac{1}{n}$$ And you probably already know that the harmonic series diverges (it can be proven by integral test).

Hint. Note that for any positive integer $n$, $$\tan\left(\frac{1}{n}\right)> \frac{1}{n}.$$ See Why $x<\tan{x}$ while $0<x<\frac{\pi}{2}$?

Note that $$\tan \left(\frac{1}{n}\right)\sim_{_{\infty}} \frac{1}{n}$$ As $$\tan \left(\frac{1}{n}\right)=\frac{1}{n}+O\left(n^{-3}\right)$$ Which can be derived from the Maclaurin series expansion of $\tan x$.