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.
 A: 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.
