I'm studying almost-surely convergence and convergence in probability of $S_{n}=X_{1}+\cdots+X_{n},$ where $X_{n}$ is distributed $\mathrm{ Poisson}\left(1/n\right)$ with $n\in\mathbb{N}$ and the sequence of $X_{n}$ are independent.
I'm stuck with this because I cannot applied Chebyshev inequality because of divergence of $\displaystyle\sum_{i=1}^{\infty}\frac{1}{n}$, neither Borel-Cantelli Lemma because I don't find a useful bound to use it.
I tried use the convergence of $\displaystyle\sum_{n=1}^{\infty}E\left(X_n\right)<\infty$ implies $\displaystyle\sum_{n=1}^{\infty}X_n$ convergence a.s., but I don't get it.
Any kind of help is very thanked.
Best Answer
Let $a_n:=\Pr\left(S_{2n}-S_n\geqslant 1\right)$. Using independence, we can see that $S_{2n}-S_n$ is Poisson distributed with parameter $\sum_{i=n+1}^{2n}1/i$. As a consequence, we have $$a_n\geqslant 1-\exp\left(-\sum_{i=n+1}^{2n}1/i\right) \geqslant 1-\exp\left(-\frac 12\right)\gt 0.$$
Therefore, the sequence $\left(S_n\right)_{n\geqslant 1}$ cannot be convergent in probability.