For what values of $a$ does $\prod_{j=1}^nX_j$ converge in probability to zero? Let $(X_n)_{n>1}$ be sequence of independent random variables such that $X_n \sim Bern(1-\frac{1}{n^a})$ where $a>0$ is a constant. Let $Y_n=\prod_{j=1}^n X_j.$ For what values of $a$ does $Y_n$ converge in probability to zero?
My approach is the following:
$P(Y_n>0)=P((X_1,\ldots,X_n)>0)=\prod_j P(X_j>0)=\prod_j P(X_j=1)=(1−1/n^a)^$. To ensure convergence in probability, we require $P(Y_n>0)\to 0$ as $n \to \infty$. So it is required to check $\lim_{n\rightarrow \infty}(1−1/n^a)^n = 0$.
I am stuck here. I know that $\lim_{n\to \infty}(1−1/n)^n=e^{−1}$ but there is the $n^a$ in the denominator so it is not clear how to use the limiting formula. Any help is appreciated.
 A: Hint:
$$\lim_{n\rightarrow \infty}(1−\frac{1}{n^a})^n 
=\lim_{n\rightarrow \infty}(1−\frac{1}{n^a})^{n^{1-a+a}}  $$
$$=\lim_{n\rightarrow \infty}(1−\frac{1}{n^a})^{n^{+a}n^{1-a}}  $$
$$=\lim_{n\rightarrow \infty}\left((1−\frac{1}{n^a})^{n^{a}}\right)^{n^{1-a}}  $$
$$=\lim_{n\rightarrow \infty} \left(e^{-1}\right)^{n^{1-a}}  $$
$$=\lim_{n\rightarrow \infty} \left(e^{-n^{1-a}}\right)  $$
for $1-a>0$ , $n^{1-a}\rightarrow +\infty$ so $\left(e^{-n^{1-a}}\right) \rightarrow 0$
for $1-a=0$ , $n^{1-a}\rightarrow ?$ 
for $1-a<0$ , $n^{1-a}\rightarrow ?$
A: My approach is the following:
$P(Y_n>0) = P((X_1,....,X_n)>0) = \Pi_j P(X_j>0) = \Pi_j P(X_j=1) = (1-\frac{1}{n^a})^n$. To ensure convergence in probability, we require $P(Y_n>0) \rightarrow 0, as n \rightarrow \infty$. So it is required to check $lim_{n-\rightarrow \infty}(1-\frac{1}{n^a})^n = 0.$ 
Following the hints provided, the convergence is ensured when $a>1$. 
Now what about $L^2$ convergence? For that we require $lim_{n-\rightarrow \infty} E[Y_n^2] =0$. This reduces to finding $a$ such that
$$lim_{n-\rightarrow \infty} \Big(\frac{1}{n^a}\big(1-\frac{1}{n^a}\big)\Big)^n =0$$ This looks trickier. 
A: You’ve shown that convergence in probability to zero fails for $a\geq 1$. Use the inequality $1-x\leq \exp(-x)$ for the case of $a<1$. 
