Let $X_n$ be independent Poisson random variables with $\mathbb{E}[X_n] = \lambda_n$. Define $S_n = X_1 + \dots + X_n$. Show that if $\sum \lambda_n = +\infty$, then $\frac{S_n}{\mathbb{E}[S_n]} \longrightarrow 1$ a.s.
How to prove this?
My attempt is to show, first, that $\frac{S_n}{\mathbb{E}[S_n]} \longrightarrow 1$ in probability and then show that $\frac{S_n}{\mathbb{E}[S_n]} \longrightarrow 1$ a.s.
Here is a solution that uses SLLN and the additive property of the Poisson distribution.
Let $(N_t)_{t \geq 0}$ be a Poisson process of unit rate. Then by the SLLN, together with the inequality
$$ \frac{N_{[t]}}{[t]+1} \leq \frac{N_t}{t} \leq \frac{N_{[t]+1}}{[t]}, $$
it is easy to check that $N_t / t \to 1$ as $t \to \infty$ a.s. Now let $T_k = \lambda_1 + \cdots + \lambda_k$. Then $T_k \to \infty$ and hence $ N_{T_k} / T_k \to 1$.
Finally, notice that the distribution of the process $(S_n : n \geq 1)$ is identical to the distribution of the process $(N_{T_k} : k \geq 1)$. Therefore $S_n / \Bbb{E}S_n \to 1$ as $n \to \infty$ a.s.