
Originally Posted by
polymerase
Does:
$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{\tan\left(\f rac{1}{n}\right)}{n}$ converge?
Yes, it converges.
Note that,
$\displaystyle \frac{\tan \frac{1}{n}}{n} = \frac{\sin \frac{1}{n}}{n\cos \frac{1}{n}} $
Furthermore,
$\displaystyle \cos \frac{1}{n} \geq \kappa$ for all $\displaystyle n\geq 1$, where $\displaystyle \kappa > 0$
Thus,
$\displaystyle \frac{\sin \frac{1}{n}}{n\cos \frac{1}{n}} \leq \frac{\sin \frac{1}{n}}{\kappa \cdot n} \leq \frac{\frac{1}{n}}{\kappa \cdot n} = \frac{1}{\kappa \cdot n^2}$
And the above series converges, now apply direct comparison.