Prove the random variable is bounded Let $g_n$ be given by
$$g_n=\prod_{i=1}^n \left(1+\frac{X_i}{\sqrt{i}}\right)$$
where $X_i$'s are independent random variables with $P(X_i=1)=P(X_i=-1)=0.5$.Then is it true that $P(g_n \rightarrow \infty)=0$
.If yes how can I show it?
My attempt:I think I have to use Borel Cantelli lemma but I am not sure how.
 A: In fact, a much stronger result holds: $P[g_n \to 0]=1$. This is an application of Kakutani's theorem for product Martingales, see Probability with Martingales by David Williams, Section 14.12.
Back to your problem, let $Y_i:= 1+\frac{X_i}{\sqrt{i}}$, and $a_i:=\mathbb{E}\left[\sqrt{Y_i}\right]$. Based on Kakutani's theorem, to show that $P[g_n \to 0]=1$, it suffices to prove that $\sum_{i=1}^\infty(1-a_i)=\infty$. I will now show this.
With an easy calculation, $a_i=\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{i}}}+\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{i}}}$. I will show that $a_i=1-\Theta\left(\frac{1}{i}\right)$, which finishes the proof.
I claim that for any $x,y\in [0,2]$, and for some absolute constant $c>0$, we have $\frac{1}{2}\sqrt{x}+\frac{1}{2}\sqrt{y}+c(x-y)^2\leq \sqrt{\frac{x+y}{2}}$. This is because the function $f(x)=\sqrt{x}$ is strongly-concave inside $[0,2]$. Setting $x=1+\frac{1}{\sqrt{i}}$ and $y=1-\frac{1}{\sqrt{i}}$, we are done.
A: For each $n$, define $M_n=\prod^n_{j=1}\big(1+\tfrac{X_j}{\sqrt{j}}\big)$, and set $\mathcal{F}_n=\sigma(X_1,\ldots,X_n)$.
Observe that $M_n\geq0$ for each $n\in\mathbb{N}$. Since
$E\Big[1+\frac{X_n}{\sqrt{n}}\Big]=\frac12(1-\frac{1}{\sqrt{n}})+ \frac12(1+\frac{1}{\sqrt{n}})=1$ for all $n$, the independence of the sequence $\{X_n\}$ implies that $M_n\in\mathcal{L}_1(\mathbb{P})$ for each $n$ and
$$E[M_{n+1}|\mathcal{F}_n]=M_n E\big[(1+\frac{X_{n+1}}{\sqrt{n+1}}\big)\big]=M_n$$
Hence $\{M_n,\mathcal{F}_n\}$ is a martingale bounded in $L_1(\mathbb{P})$. By the martingale convergence theorem $M_n\xrightarrow{n\rightarrow\infty}M$ $\mathbb{P}$-a.s. for some $M\in L_1(\mathbb{P})$.
Furthermore,
$$E[M_n]=E\Big[\prod^n_{j=1}\big(1+\tfrac{X_j}{\sqrt{j}}\big)\Big]=\prod^n_{j=1}E\Big[\big(1+\tfrac{X_j}{\sqrt{j}}\big)\Big]=1$$
By Fatou's lemma $$E[M]=E\Big[\prod^\infty_{j=1}\big(1+\tfrac{X_j}{\sqrt{j}}\big)\Big]\leq \liminf_nE\Big[\prod^n_{j=1}\big(1+\tfrac{X_j}{\sqrt{j}}\big)\Big]=1<\infty$$
