Discuss Convergence of Series $n (e^{-1/n^2} - \cos^2{(\frac{1}{n}})) $

Question

I presume this series converges to 0, how do I prove it?

$$\require{cancel} n (e^{-1/n^2} - \cos^2{(\frac{1}{n}})) $$ If I do as @Alex suggested i get \begin{align*} \sim &n(1-\frac{1}{n^2}+\frac{1}{2n^4} -(1-\frac{1}{2n^2})^2)\\ &=n(\cancel{1}-\cancel{\frac{1}{n^2}}+\frac{1}{2n^4}-\cancel{1}+\cancel{\frac{1}{n^2}}-\frac{1}{4n^4})\\ &=\frac{n}{4}(\frac{2}{n^4}-\frac{1}{n^4}) = \frac{1}{4n^3}\stackrel{\infty}\to0 \end{align*}

Answer 1

Expand $e^{-\frac{1}{n^2}} \sim 1- \frac{1}{n^2} + \frac{1}{2n^4}$ and $\cos \frac{1}{n} \sim 1-\frac{1}{2n^2}$.