Suppose $\{X_n\}_{n=1}^{+\infty}$ is a sequence of independent variables.
I'm trying to prove that if $\forall n \in \mathbb{N}: \mathbb{E}X_n = 0$ and $\exists C\in \mathbb{R}: \sup\limits_{n \in \mathbb{N}} \mathbb{E}|X_n|^4 \leq C$, then $\frac{1}{n} (X_1+X_2+\dots+X_n) \to 0$ almost surely.
First, my question is, is that true that the second premise $\sup\limits_{n \in \mathbb{N}} \mathbb{E}|X_n|^4 \leq C$ is equivalent to this: $\forall n \in \mathbb{N}: \mathbb{E}|X_n|^4 \leq C$?
I was trying to apply Kolmogorov's strong law. For this, I have to prove that this series converges: $$\sum\limits_{n=1}^{+\infty} \dfrac{\text{Var}(X_n)}{n^2}$$
How can I approach this problem? I would appreciate any help.
Kolmogorov's strong law is overkill in this scenario. Let $S_n=\sum_{i=1}^nX_i$. Then, since $\mathbb E[X_i] = 0$ for all $i$,
$$\mathbb E[S_n^4] = \sum_{i=1}^n\mathbb E[X_i^4] + 2\sum_{1\le i<j\le n}\mathbb E[X_i^2]\mathbb E[X_j^2] \le nC + n(n-1)C = n^2C,$$
where the Cauchy-Schwarz inequality is used to bound the second sum. Hence using Chebyshev's inequality, for any $\varepsilon>0$ one has
$$ \mathbb P\left(\left|\frac1nS_n\right|>\varepsilon\right) \le \frac{\mathbb E[S_n^4]}{n^4\epsilon^4} \le \frac{C}{n^2\epsilon^4}. $$
It follows that $\sum_{n=1}^\infty \mathbb P\left(\left|\frac1nS_n\right|>\varepsilon\right) < \infty$, and so by Borel-Cantelli we have that $\left|\frac1nS_n\right|>\varepsilon$ only finitely many times, with probability one. Since this is true for any $\varepsilon>0$, it follows that $\frac1nS_n\to0$ almost surely.