Let $(X_i)_{i \in \mathbb{N}} \subset \mathcal{L}^2$ be independent and $(a_n)_{n \in \mathbb{N}} \subset \mathbb{R}_{\ge 0}$ an increasing an unbounded sequence such that
$$\sum_{i=1}^{\infty} \frac{\mathbb{V}[X_i]}{{a_i}^2} < \infty $$
I want to use Doob’s convergence theorem to show that
$$\lim_{n \to \infty} \frac{1}{a_n}\sum_{i=1}^{n}\big(X_i - \mathbb{E}[X_i]\big) = 0 \textit{ a.s.} $$
My problem is that while I can show that $R_n := \frac{1}{a_n}\sum_{i=1}^{n}\big(X_i - \mathbb{E}[X_i]\big)$ is a supermartingale and $\operatorname{sup}_{n \in \mathbb{N}} \mathbb{E}\big[R_n\big] < \infty$, Doob's convergence theorem doesn't give me any information about the limit itself.
Assume without loss of generality that $X_i$ is centered. Let $Y_i:=\sum_{k=1}^iX_k/a_k$ for $i\geqslant 1$ and $Y_0=0$. Then $$ \frac{1}{a_n}\sum_{i=1}^nX_i=\frac{1}{a_n}\sum_{i=1}^na_i\left(Y_i-Y_{i-1}\right) =Y_n+\frac{1}{a_n}\sum_{i=1}^{n-1}\left(a_i-a_{i+1}\right)Y_i.$$ From the assumption, $(Y_i)_{i\geqslant 1}$ is a martingale which is bounded in $\mathbb L^2$. Therefore, there exists a random variable $Y$ such that $Y_i\to Y$ almost surely. From the assumptions on $(a_i)$, we can see that $\frac{1}{a_n}\sum_{i=1}^{n-1}\left(a_i-a_{i+1}\right)Y_i\to -Y$ almost surely.