Find the almost sure limit of $X_n/n$, where each random variable $X_n$ has a Poisson distribution with parameter $n$ 
$X_{n}$ independent and $X_n \sim \mathcal{P}(n) $  meaning that $X_{n}$ has Poisson distributions with parameter $n$. What is the $\lim\limits_{n\to \infty} \frac{X_{n}}{n}$ almost surely ? 

I think we can write $X(n) \sim X(1)+X(1)+\cdots+X(1)$ where the sum is taken on independent identical distribution then use the law of large number. But I am not sure that is it correct or not. Can anyone give me some hints? Thank you in advance!
 A: If $X_n\sim\mathsf{Pois}(n)$ for $n=1,2,\ldots$, then $n^{-1}X_n\stackrel{\mathrm{a.s.}}\longrightarrow 1$. It suffices to show that for all $\varepsilon>0$, 
$$\mathbb P\left(\liminf_{n\to\infty}\left|n^{-1}X_n-1\right|<\varepsilon\right)=1. $$
We have $$\left\{\left|n^{-1}X_n-1\right|<\varepsilon\right\}^c = \left\{X_n\leqslant n(1-\varepsilon) \right\}\cup\left\{X_n\geqslant n(1+\varepsilon) \right\}, $$
so the Chernoff bounds yield $$\mathbb P(X_n\leqslant n(1-\varepsilon)) \leqslant \frac{e^{-n}(ne)^{n(1-\varepsilon)}}{(n(1-\varepsilon))^{n(1-\varepsilon)}}=  \left(e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon} \right)^{-n} $$
and
$$\mathbb P(X_n\geqslant n(1+\varepsilon)) \leqslant \frac{e^{-n}(ne)^{n(1+\varepsilon)}}{(n(1+\varepsilon))^{n(1+\varepsilon)}}=
\left(e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} \right)^{-n} . $$ 
Now
$$e^\varepsilon(1-\varepsilon)^{1-\varepsilon} = e^{\varepsilon + (1-\varepsilon)\log(1-\varepsilon)}=e^{\frac{\varepsilon^2}2+O(\varepsilon^3)}>1 $$
and 
$$e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} = e^{-\varepsilon+(1+\varepsilon)\log(1+\varepsilon)}=e^{\frac{\varepsilon^2}2+O(\varepsilon^3)}>1, $$
so $$\sum_{n=1}^\infty \mathbb P(X_n\leqslant n(1-\varepsilon))\leqslant\sum_{n=1}^\infty \left(e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon} \right)^{-n}<\infty $$ and 
$$\sum_{n=1}^\infty \mathbb P(X_n\geqslant n(1+\varepsilon))\leqslant\sum_{n=1}^\infty \left(e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} \right)^{-n}<\infty.$$
It follows from the first Borel-Cantelli lemma that $$\mathbb P\left(\limsup_{n\to\infty} \{X_n\leqslant n(1-\varepsilon)\}\right)=\mathbb P\left(\limsup_{n\to\infty} \{X_n\geqslant n(1+\varepsilon)\}\right)=0,$$
and so we conclude.
