Convergence of an infinite sum of weighted independent and identically distributed random variable Let $z_i$ be $i.i.d$ random variable with $E(z_i)=0$ and $E(Z_i^2)=1$ with a symmetric distribution. Further, $|\beta|<1$.
Now consider $\sum\limits_{i=1}^{\infty} \beta^i (z_i+|z_i|)$. I want to prove this summation converges almost surely ($a.s$).
End goal is finding $E(exp(\sum\limits_{i=1}^{\infty} \beta^i (z_i+|z_i|)))$. So if that sum converges almost surely I can move the expectation inside the infinite product as $\prod_{i=1}^{\infty} E(exp(\beta^i (z_i+|z_i|)))$.

I believe that this can be handled by Borel-Cantelli's lemma : if you let $\alpha>1$ you have for any integer $k$ the inequality
$$\begin{align}
\mathbb P\left(|\beta^k(z_k+|z_k|)|>\frac{1}{k^\alpha}\right) &=\mathbb P\left(|z_k+|z_k||>\frac{1}{\beta^kk^\alpha}\right)\\
&=\mathbb P\left(z_k+|z_k|>\frac{1}{\beta^kk^\alpha}\right)\\
&+ \mathbb P\left(z_k+|z_k|<-\frac{1}{\beta^kk^\alpha}\right)\\
&=\mathbb P\left(z_k+|z_k|>\frac{1}{\beta^kk^\alpha}\right) + 0\tag1\\
&\le \beta^kk^\alpha\mathbb E[z_k + |z_k|]\tag2\\
&=\beta^kk^\alpha\mathbb E[|z_k|] = \beta^kk^\alpha\mathbb E\left[\sqrt{z_k^2}\right] \\
&\le\beta^kk^\alpha \sqrt{\mathbb E\left[z_k^2\right]}=\beta^kk^\alpha\tag3\end{align}$$
Where I used in $(1)$ the fact that $z_k+|z_k|$ is almost surely non-negative, Markov's inequality in $(2)$ and Jensen's inequality in $(3)$.
Now, because $|\beta|<1$, if we denote by $E_k$ the event $\{|\beta^k(z_k+|z_k|)|>1/k^\alpha\} $, we have just proven that
$$ \sum_{k=1}^\infty \mathbb P(E_k) <\infty$$
Which implies by Borel Cantelli that the event $\{E_k \text{ happens infinitely often}\} $ has probability zero. In other words, this means that there exists a $k_0$ such that
$$\mathbb P\left(|\beta^k(z_k+|z_k|)|\le\frac{1}{k^\alpha}\right) = 1 \text{ for all } k\ge k_0 $$
We can finally conclude that, almost surely
$$\begin{align}\sum_{k=1}^\infty|\beta^k(z_k+|z_k|)| &= \sum_{k=1}^{k_0-1}|\beta^k(z_k+|z_k|)| + \sum_{k=k_0}^\infty|\beta^k(z_k+|z_k|)|\\
&\le \sum_{k=1}^{k_0-1}|\beta^k(z_k+|z_k|)| + \sum_{k=k_0}^\infty\frac{1}{k^\alpha}\\
&<\infty\end{align}$$
As desired.
