Almost sure convergence of a sum of independent random variables Let $\{X_n\}_{n\geq1}$ be a sequence of centered independent random variables such as $E(X_n^2)=2n$
and $$Y_n=\frac1{n^\alpha}\sum_{i=1}^{i=n}X_i\quad\quad\alpha\geq1$$
I am trying to prove that for $\alpha > \frac32$, $Y_n \rightarrow 0$ almost surely.

I started by calculating $Var(Y_n)$ which I found to be equal to $\frac{n(n+1)}{n^{2\alpha-2}}$ and since $E(Y_n)=0$ we get
$$\lim_{n\to\infty} E(|Y_n-0|^2)=0$$
which means that for $\alpha > 1$,  $\{Y_n\}_{n\geq1}$ converges to $0$ in quadratic mean.
I am stuck going form there to the almost sure convergence.
Any help would be greatly appreciated!

Let us recall:
(1): Kronecker's Lemma: If $(x_n)_{n \in \mathbb N}$ is a sequence such that $\sum_{n} x_n $ converges, $(b_n)_{n \in \mathbb N}$ is increasing positive sequence such that $\lim_{n \to \infty} b_n = +\infty$, then $\lim_{n \to \infty} \frac{1}{b_n} \sum_{j=1}^n x_jb_j = 0$
(2): Kolmogorov's two-series theorem: If $(X_n)_{n \in \mathbb N}$ are independent, $\mathbb E[X_n], Var(X_n)$ exists and are finite for every $n \in \mathbb N$, series $\sum \mathbb E[X_n] , \sum Var(X_n)$ are convergent, then $\sum X_n$ converges almost surely.
Using those above, we'll prove:
Lemma: Let $(X_n)_{n \in \mathbb N}$ be independent rvs such that $Var(X_n)$ is finite for every $n \in \mathbb N$. Moreover, let $(b_n)_{n \in \mathbb N}$ be increasing sequence of positive numbers with $\lim b_n = \infty$. Let $S_n = \sum_{j=1}^n X_n$. If $\sum \frac{Var(X_n)}{b_n^2}$ converges then $\frac{S_n - \mathbb E[S_n]}{b_n}$ converges to $0$ almost surely.
Proof: Let $Y_j = \frac{X_j - \mathbb E[X_j]}{b_j}$, then for every $j \in \mathbb N$ we have: $\mathbb E[Y_j] = 0, Var(Y_j) = \frac{Var(X_j)}{b_j^2}$, so both $\sum \mathbb E[Y_j], \sum Var(Y_j)$ converges, so using (2): we get $\sum Y_j$ converges almost surely. So we have a set of $\mathbb P$ measure $1$ where for every $\omega$ in that set, we can use (1) with sequence $x_n = Y_n(\omega)$ to obtain: $\lim_{n \to \infty} \frac{1}{b_n} \sum_{j=1}^n Y_j(\omega)b_j = 0$
But $Y_j(\omega)b_j = X_j(\omega) - \mathbb E[X_j]$, so $\sum_{j=1}^n Y_j(\omega)b_j = S_n(\omega) - \mathbb E[S_n]$. So on the set of measure $1$ we have that convergence, so $\frac{S_n - \mathbb E[S_n]}{b_n}$ converges almost surely to $0$.
Answer: You can just use this with $S_n = \sum_{j=1}^n X_j$, then $\mathbb E[S_n] = 0$. Moreover $Var(X_n) = 2n$, so $\sum \frac{Var(X_n)}{n^{2a}} $ converges iff $2a-1 > 1$ which means $a>1$. So even for $a>1$ it holds that $Y_n = \frac{S_n - \mathbb E[S_n]}{n^a}$ converges almost surely to $0$.
