Let $\{X_r\}_{r\ge1}$ be independent Poisson random variables with respective parameters $\{\lambda_r\}_{r\ge1}$. Show that $\sum_\limits{r\ge1} X_r$ converges or diverges almost surely according as $\sum_\limits{r\ge1} \lambda_r$ converges or diverges.
I know this question has already been asked and has an answer outlined here but I have a question on the solution and given that the post is so old, I thought I would make a new post with a partial solution then pose my question.
Since Poisson random variables take values on the set $\{0,1,2,3,..\}$ we have that:
\begin{align}
\sum_\limits{r\ge1} X_r < \infty \ \text{a.s}
&\iff \mathbb{P}(X_r=0\ \ \text{eventually})=1\\
&\iff \mathbb{P}(X_r>0\ \ \text{i.o.})=0\\
&\iff \sum_\limits{r\ge1} \mathbb{P}(X_r>0)<\infty \quad \text{by Borell-Cantelli Lemmas}
\end{align}
Now note that:
\begin{align}
\sum_\limits{r\ge1} \mathbb{P}(X_r>0)=\sum_\limits{r\ge1} (1- \mathbb{P}(X_r=0))=\sum_\limits{r\ge1} (1-e^{-\lambda_r})\le\sum_\limits{r\ge1} \lambda_r
\end{align}
Thus, $\sum_\limits{r\ge1} \lambda_r < \infty \implies \sum_\limits{r\ge1}\mathbb{P}(X_r > 0)<\infty \implies \sum_\limits{r\ge1} X_r < \infty \ \text{a.s}$, and half the result is proved.
Now, again since Poisson random variables take values on the set $\{0,1,2,3,..\}$ we have that:
\begin{align}
\sum_\limits{r\ge1} X_r = \infty \ \text{a.s}
&\iff \mathbb{P}(X_r=0\ \ \text{eventually})=0\\
&\iff \mathbb{P}(X_r>0\ \ \text{i.o.})=1\\
&\iff \sum_\limits{r\ge1} \mathbb{P}(X_r>0)= \infty \quad \text{by Borell-Cantelli Lemmas}
\end{align}
Now note that:
\begin{align}
\sum_\limits{r\ge1} \mathbb{P}(X_r>0)=\sum_\limits{r\ge1} (1-e^{-\lambda_r}) \ge \sum_\limits{r\ge1} \Bigl(\frac{\lambda_r}{\lambda_r +1}\Bigr)
\end{align}
Now we want to claim that: $\sum_\limits{r\ge1} \lambda_r = \infty \implies \sum_\limits{r\ge1}\mathbb{P}(X_r > 0) = \infty \implies \sum_\limits{r\ge1} X_r = \infty \ \text{a.s}$, completing the proof.
The only issue here is that this last claim is a little less obvious, so my question is does
\begin{align}
\sum_\limits{r\ge1} \lambda_r = \infty \implies \sum_\limits{r\ge1} \Bigl(\frac{\lambda_r}{\lambda_r +1}\Bigr)= \infty?
\end{align}
If so, is this an easy thing to see and I am just missing it or does it require a bit of unpacking? Or is there a better way to show the second half of the result?
Best Answer
If $\lambda \le 1$, then $\lambda/(1+\lambda) \ge \lambda/2$, while if $\lambda > 1$, then $\lambda/(1+\lambda) > 1/2$. If infinitely many $\lambda_i > 1$, then infinitely many $\lambda_i/(1+\lambda_i) > 1/2$ and the second sum diverges. If not, use a Limit Comparison Test.