I have a sequence $X_n$ of independent r.v. such that $X_i \sim \mathcal{N}\left(0,\frac1i\right)$. Then I have that $Y_i = e^{X_i}$ and $Z_n= \prod_{j=1}^n Y_i$.
I need to give a tight, asymptotic upper-bound on the following event: $$P(\{Z_n\geq n^c\})$$ with $c>0$ fixed. And afterwards it asks me for which values of $c>0$ we can say that $P(\{Z_n\geq n^c \text{ infinitely often}\})=0$.
I also have a hint: $\lim_{n\to\infty}\frac{1}{\log n} \sum_{j=1}^n 1/j = 1$
I tried to give some upper-bound, but I do not find the right function $\phi$ to use in order to say that $P(\{Z_n\geq n^c\}) \leq \frac{E[\phi(Z_n)]}{\phi(n^c)}$. To solve the second problem it suffices to show that $\sum_{n}P(\{Z_n\geq n^c\}) < \infty$ and by the Borel-Cantelli lemma I can say that the event happens infinitely often with probability $0$. Any hint on what function should I choose? Also, can you describe me the rationale, so that I can get better in this kind of exercises? Thanks!
Let $s_n:=\sum_{i=1}^n1/i$ and $S_n:=\sum_{i=1}^nX_i$. Then by independence, $Z_n=e^{S_n}$ and $S_n$ has a normal distribution, with mean $0$ and variance $s_n$. Therefore, $$ \Pr\left\{Z_n\geqslant n^c\right\}=\Pr\left\{\exp\left(s_n^{1/2}N\right)\geqslant n^c\right\}, $$ where $N$ has a standard normal distribution. The latter probability can be express in terms of tail of a standard normal distribution, for which asymptotics are known. Then we have to use the given equivalent for $s_n$.