Let $(X_n)_n$ be a sequence i.i.d random variable following Poisson distribution $P(\lambda).$
Prove that, almost surely, $$\lim_n\frac{\max_{1\leq k \leq n}X_k\ln(\ln(n))}{\ln(n)}=1.$$
Hint: Use Borel-Cantelli and the inequality: $$\frac{\lambda^ne^{-\lambda}}{n!} \leq P(X_1 \geq n)\leq \frac{\lambda^n}{n!}.$$
Attempt: Since we should use Borel-Cantelli, tried to prove, for $0<\epsilon<1,$ that $$\sum_nP(|\frac{\ln(\ln(n))}{\ln(n)}\max_kX_k-1|>\epsilon)<+\infty.$$ So first, letting $x_n=\left \lfloor{(\epsilon+1)\frac{\ln(n)}{\ln(\ln(n))}}\right \rfloor,$ we should prove that $\sum_n(1-(1-P(X_1 \geq x_n+1))^n)$ converges but since $$1-(1-P(X_1 \geq x_n+1))^n \leq 1-(1-\frac{\lambda^{x_n+1}}{(x_n+1)!})^n,$$ so we have to prove that $\sum_n(1-(1-\frac{\lambda^{x_n+1}}{(x_n+1)!})^n)<\infty,$ but $1-(1-\frac{\lambda^{x_n+1}}{(x_n+1)!})^n$ is equivalent to $n\lambda^{x_n+1}\frac{1}{(x_n+1)!}$
How to prove that $\sum_n n\lambda^{x_n+1}\frac{1}{(x_n+1)!}$ converges? Is it possible to prove the problem without probability theory?