# Limit superior of iid Poisson random variables

Im trying to undestand the proof of the next: Suppose $$X_n$$ are iid Poisson random variable with rate $$\lambda>0.$$ Prove that $$\displaystyle\limsup\frac{X_n\log(\log n)}{\log(n)}=1\space a.s.$$

The proof given is following: We have $$P(X\geq n)\leq e^{\lambda}P(X=n).$$

Let $$a_n$$ be the integer part of $${\delta \log n \over \log\log n}$$ for a $$\delta >0$$. Then

$$P(X=a_n)={e^{-\lambda}\lambda^{a_n}\over a_n!}=e^{-\lambda}e^{a_n \log \lambda}e^{-\sum_{j=1}^{a_n}\log j}=e^{-a_n\log a_n (1+o(1))}=n^{-\delta +o(1)}$$

Then, $$\sum_{n=1}^{\infty} P(X_n\geq a_n)<\infty$$ or $$=\infty$$ depending upon $$\delta >1$$ or $$\delta<1$$. Borel-Cantelli lemma finishes the proof.

I'm stuck in this: $$e^{-\lambda}e^{a_n \log \lambda}e^{-\sum_{j=1}^{a_n}\log j}=e^{-a_n\log a_n (1+o(1))}=n^{-\delta +o(1)}.$$ I don't get such expressions; I have understood that little o-notation means a quotient between two functions have zero limit but I don't know how this was used here to get such expressions.

Any kind of help is thanked in advanced.

The notation $$b_n=o(c_n)$$ means $$b_n/c_n\to0$$ as $$n\to\infty$$. Therefore, $$b_n=o(1)$$ simply means $$b_n\to0$$. Thus, the first equality is asserting that $$-\lambda + a_n\log\lambda - \sum_{j=1}^{a_n}\log j = -a_n\log a_n(1 + b_n),$$ for some sequence $$\{b_n\}$$ satisfying $$b_n\to0$$. This is equivalent to $$\frac{\lambda - a_n\log\lambda + \sum_{j=1}^{a_n}\log j}{a_n\log a_n} \to 1.$$ The main thing needed to prove this is to show that $$\sum_{j=1}^n \log j\sim n\log n$$, meaning their ratio tends to $$1$$ as $$n\to\infty$$. You can guess at this result by comparing the sum to the corresponding integral $$\int_1^n\log x\,dx$$. To check it rigorously, you could use the Stolz–Cesàro theorem.
The second equality asserts that if $$b_n\to0$$, then $$-a_n\log a_n(1 + b_n) = (-\delta + c_n)\log n$$ for some sequence $$\{c_n\}$$ satisfying $$c_n\to0$$. This is equivalent to $$\begin{equation}\label{1} \frac{a_n\log a_n}{\log n} \to \delta.\tag{1} \end{equation}$$ A complete, rigorous proof of this would need to account for the fact that $$a_n$$ is an integer. But in this sketch, I will work as though $$a_n = \frac{\delta\log n}{\log\log n}.$$ In this case, $$\log a_n = \log\delta + \log\log n - \log\log\log n.$$ The leading term here is $$\log\log n$$ and $$\frac{a_n\log\log n}{\log n} = \delta,$$ suggesting that \eqref{1} holds and verifying the second equality.