Convergence of a sequence of r.v. in probability and almost surely

Question

Let $(X_k)_{k=1}^{\infty}$ be a sequence of independent r.v. where $\mathbb{E}X_k=0$ and $\mathbb{E}X_k^2=k$. Consider $Y_n=\frac{1}{n^2}\sum_{k=1}^{n}X_k$. Show $Y_n$ converges to 0 in probability and almost surely.

We note by Strong Law, $\frac{1}{n}\sum_{k=1}^{n}X_k$ converges to 0 a.s. Thus, $Y_n$ converges a.s. Since $Y_n$ converges a.s. we have convergence in probability. Is it okay?

Answer 1

$EY_n^{2}=\operatorname{var} (Y_n)=\frac 1 {n^{4}} \sum_{k=1}^{n} \operatorname{var} (X_k)=\frac 1 {n^{4}} \sum_{k=1}^{n} k=\frac 1 {n^{4}} \frac {n(n+1)} 2$. Hence $\sum_{n=1}^{\infty} EY_n^{2} <\infty$. This implies $\sum_{n=1}^{\infty} Y_n^{2} <\infty$ almost surely and hence $Y_n \to 0$ almost surely.