What convergence test can I use on this series?

Question

I am doing practice problems for an exam, and I am not sure how to test this series:

$\lim_{n\to\infty} \cos(n)\sin^2(\frac{1}{n})$

I am supposed to use $\lim_{x\to0} \frac{\sin(x)}{x}=1$ to find the answer somehow, but I don't see how this fits since x is going to 0 instead of infinity.

Answer 1

$1/n$ goes to $0$ as $n$ goes to infinity, hence $$\lim_{n\to \infty}n^2\sin^2\left(\frac 1n\right)=1.$$ Since $$\left|\cos n\sin^2\left(\frac 1n\right)\right|\leqslant \sin^2\left(\frac 1n\right)$$ and $\sum_{n\geqslant 1}\frac 1{n^2}$ is convergent, we are done.