$$\sum_{n=17}^{\infty}\left(\tan\left(\frac{1}{n}\right)\right)^2 \ \ $$
My first guess is to write the series as integral. And use the substitution for u=1/n. That changes my upper and lower bounded limits to - lower = 1/17 and upper to 0. In that case switch the limits and put a negative sign. I fail to understand what is really going on after that.
You can use the fact that $\tan(x)$ is equivalent to $x$ when $x$ goes to $0$. Then the general term of your serie is equivalent to $\frac{1}{n^2}$. Then you can conclude that your serie is convergent by the comparison theorem, as $\sum_n \frac{1}{n^2}$ converges.