Test the series for convergence/divergence.

Question

zenmo · Answer

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

tkhunny · Answer

Do the individual terms approach zero?  That's the first thing to know.

zenmo · Answer

tan(1/n) reaches 0, since tan(1/oo) = tan(0) = 0. If looking at the whole series as a whole, 0/constant will give a sum of 0. That is my take.

holsteremission · Answer

Seems that you grasp that $\lim\limits_{n\to\infty}\dfrac{\tan\frac1n}n=0$, but just because the terms in the series approach 0 does not mean the sum itself approaches 0. All you know is that the limit test for divergence fails.

tkhunny · Answer

I wonder why i mentioned "the first thing to know"...
Second thing, what is your impression?

Unlike 1/n, which we know diverges, this series has two parts heading for zero.  That might make a difference.

holsteremission · Answer

If you're equipped with the knowledge, you could use the series expansion for $\tan x$ and come up with an appropriate series comparison.
$$\tan x=x+\frac{x^3}{5}+\frac{2x^5}{15}+\cdots$$