Does someone know the proof of the following statement:
Given a sequence of independent random variables $(Y_n)_{n=1}^{\infty}$ and a sequence of positive constants $(a_n)_{n=1}^{\infty}$ such that $a_n\to\infty$ and $$\sum_{n=1}^{\infty}\frac{\text{var}(Y_n)}{a_n^2}<\infty$$ then $$\frac{1}{a_n}\left(\sum_{i=1}^{n}Y_i - \mathbb{E}\left[\sum_{i=1}^{n}Y_i\right]\right)\rightarrow 0\text{ almost surely}$$
I have come up with the following: by the triangular inequality, $$\left|\frac{1}{a_n}\sum_{i=1}^{n}Y_i-\mathbb{E}[Y_i]\right|\le\frac{1}{a_n}\sum_{i=1}^{n}\left| Y_i-\mathbb{E}[Y_i]\right|$$
so that $$\mathbb{P}\left(\frac{1}{a_n}\sum_{i=1}^{n}\left| Y_i-\mathbb{E}[Y_i]\right|<\varepsilon\right)\le\mathbb{P}\left(\left|\frac{1}{a_n}\sum_{i=1}^{n}Y_i-\mathbb{E}[Y_i]\right|<\varepsilon\right)$$ now, for $1\le i\le n$, let $S_i^n$ be the event $$|Y_i-\mathbb{E}[Y_i]|<\frac{a_n\varepsilon}{n}$$ so that $$\mathbb{P}(S_1^n\wedge S_2^n\wedge\dots\wedge S_n^n)\le \mathbb{P}\left(\frac{1}{a_n}\sum_{i=1}^{n}\left| Y_i-\mathbb{E}[Y_i]\right|<\varepsilon\right)$$ But $X_i = |Y_i-\mathbb{E}[Y_i]|$ are indepentent random variables because the $Y_i$ are independent, so $$\mathbb{P}(S_1^n\wedge S_2^n\wedge\dots\wedge S_n^n)=\prod_{i=1}^{n}\mathbb{P}(S_i^n)=$$ $$=\prod_{i=1}^{n}\mathbb{P}\left(|Y_i-\mathbb{E}[Y_i]|<\frac{a_n\varepsilon}{n}\right)\ge\prod_{i=1}^{n}\left(1-\frac{n\sigma_i}{a_n\varepsilon}\right)$$ where $\sigma_i$ is the standard deviation of $Y_i$ and the last inequality follows from Chebyshev's inequality.
This doesn't work, but I suspect choosing the right $\varepsilon$ and bounds for $S_i^n$ will work. Any clue how? Also, are the $X_i$ really independent? How can this be proven?
Thanks in advance!
As @Botnakov N. commented, this is Kolmogorov's SLLN. The proof is rather complicated and needs some preparation work.
Hint: Suppose $X_1,X_2,\dots$ are independent r.v.s with finite expectation. Prove that if $\sum_i \operatorname{Var}(X_i)<\infty$, then $\sum_i [X_i-E(X_i)]$ converges a.s. (The proof is also hard and needs some lemmas.)
Then the assumption of the theorem would imply $$ \sum_i \frac{Y_i-E(Y_i)}{a_i} $$ converges a.s. The conclusion thus holds by a result in elementary calculus known as Kronecker's Lemma.
You would probably need a book to check the proofs there.