Application of Borel Cantelli I'm trying to solve the question below and after an hour no success yet! Here is the question:
$X_1, X_2, ...$ are independent poisson random variables with $EX_j = \lambda_j$. Assume that $ 0 < \lambda_j < 1$. Define $S_n = \sum_{i = 1}^{n} X_i$. We need to show that:
$$ \text{if } \sum_{j = 1}^{\infty} \lambda_j = \infty \text{ then} \frac{S_n}{ES_n} \rightarrow 1 \text{ almost surely}$$.
I appreciate if you could give me some hints on how I should approach this question.
There is a hint on this question that Borel-Cantelli lemma should be used. As you know Borel-Cantelli (lemma 2) says:
Let $A_1, A_2, ...$ be events in the probability space. Then:
if $A_1, A_2, ...$ are independent and $\sum_{k = 1}^{\infty} p(A_k) = \infty$ then $\rightarrow p(A_k, \text{ infinitely often}) = 1$
Thank you for your help.
 A: First we remark that $S_n \sim \text{Poi} \left(\mu_n \right)$ where $\mu_n := \sum_{j=1}^n \lambda_j$. Moreover, $\mathbb{E}S_n = var S_n=\mu_n$.
Let $\varepsilon>0$. By applying Tschebyscheff-inequality
$$\mathbb{P}\left[ \left| \frac{S_n}{\mathbb{E}S_n} -1 \right|>\varepsilon \right] \leq \frac{1}{(\varepsilon \cdot \mathbb{E}S_n)^2} \cdot \text{var} S_n = \frac{1}{\varepsilon^2} \cdot \frac{1}{\mu_n} \to 0 \quad (n \to \infty)$$ 
since $\mu_n= \sum_{j=1}^n \lambda_j \to \infty$. Hence $\frac{S_n}{\mathbb{E}S_n} \to 1$ in probability. Now let $n_k := \inf\{n; \mathbb{E}S_n \geq k^2\}$. Then the same calculation shows
$$\mathbb{P}\left[ \left| \frac{S_{n_k}}{\mathbb{E}S_{n_k}} -1 \right|>\varepsilon \right] \leq \frac{1}{\varepsilon^2} \cdot \frac{1}{k^2} \\
\Rightarrow \sum_k \mathbb{P}\left[ \left| \frac{S_{n_k}}{\mathbb{E}S_{n_k}} -1 \right|>\varepsilon \right] \leq \sum_k \frac{1}{\varepsilon^2} \cdot \frac{1}{k^2}  < \infty$$
By applying the Lemma of Borel-Cantelli we obtain $\frac{S_{n_k}}{\mathbb{E}S_{n_k}} \to 1$  almost surely as $k \to \infty$. Since
$$\frac{S_{n_k}(w)}{\mathbb{E}S_{n_{k+1}}} \leq \frac{S_n(w)}{\mathbb{E}S_n} \leq \frac{S_{n_{k+1}}(w)}{\mathbb{E}S_{n_k}}$$
for all $n_k<n \leq n_{k+1}$ and $\frac{\mathbb{E}S_{n_{k+1}}}{\mathbb{E}S_{n_k}} \to 1$ as $k \to \infty$ we conclude $\frac{S_n}{\mathbb{E}S_n} \to 1$ almost surely.
