Show that $\sum_{k=1}^n \frac{X_k}{k^2}$ converges a.s

$$ \mathbb{P}(|X_k|\geqslant 2\log k)=\int_{2\log k}^{+\infty}e^{-t}dt=\frac{1}{k^2} $$ Thus the series $\sum\mathbb{P}(|X_k|\geqslant 2\log k)$ converges, by Borel-Cantelli lemma we have $$ \mathbb{P}\left(\bigcap_{n\geqslant 1}\bigcup_{k\geqslant n}\{|X_k|\geqslant 2\log k\}\right)=0 $$ Because of what said above, there exists a finite number of $k$ such that $|X_k|\geqslant 2\log k$ almost surely. Thus for $k\gg1$, $|X_k|<2\log k$ almost surely so that $$ \frac{X_k}{k^2}=\mathcal{O}\left(\frac{\log k} {k^2}\right)=\mathcal{O}\left(\frac{1}{k^{3/2}}\right) $$ and thus the series $\sum\frac{X_k}{k^2}$ converges almost surely.