$\displaystyle \sum^{\infty}_{k=1}\sin(\frac{1}{4k})$
$\displaystyle \sum^{\infty}_{n=1}\tan(\frac{1}{n})$
I really have no Idea where to start.
Notice that for both of these series that as n (or k) gets very large, the value 1/n (or 1/(4k)) gets very close to zero. Consider that the derivative of sin(x) and tan(x) at points very near zero is nearly equal to one, which is like saying at points very near zero these functions behave similarly to f(x)=x. So, the functions you give could be said to be comparable to f(x)=1/x.
Brian
I figured it out. You use limit comparison test for
$\displaystyle \lim_{k\rightarrow \infty }\frac{a_{k}}{b_{k}}=\lim_{k\rightarrow \infty }\frac{\sin (\frac{1}{4k})}{\frac{1}{4k}}$
then
$\displaystyle \lim_{\theta \to 0}\frac{\sin \theta }{\theta }=1> 0$
so they both diverge or converge.
And since $\displaystyle \sum_{k=1}^{\infty }\frac{1}{4k}$ diverges by the p-series test so does $\displaystyle \sum_{k=1}^{\infty }\sin \left ( \frac{1}{4k} \right )$