Convergence in probability for $\frac { \overline { X } (1-\overline { X } ) }{ \sqrt { n } -{ 1 }/{ \sqrt { n } } }$

Question

Let $X_1,\ldots,X_n \overset {\text{iid}}\sim \operatorname{Bernoulli}(p),$ and let $\overline { X } $ be the sample mean.

So I want to show how $\frac {\overline{X} (1-\overline{X})}{ \sqrt{n} - 1/{\sqrt n}}$ converges to zero.

$\frac { \overline { X } (1-\overline{X}) }{ \sqrt{n} - 1/{\sqrt n}} \overset{P}{\longrightarrow} 0$

Any hint?

Answer 1

We can use the bound for non-negative random variable $X$ and $\epsilon>0$ that $$ P(X\geq \epsilon)\leq \frac{1}{\epsilon}E[X]. $$ Here definitely we have $\frac{\bar{X}(1-\bar{X})}{\sqrt{n}-1/\sqrt{n}}\geq 0$, so $$ P(\frac{\bar{X}(1-\bar{X})}{\sqrt{n}-1/\sqrt{n}}\geq \epsilon)\leq \frac{1}{\epsilon}E[\frac{\bar{X}(1-\bar{X})}{\sqrt{n}-1/\sqrt{n}}] \leq \frac{1}{\epsilon}\frac{p}{\sqrt{n}-1/\sqrt{n}}\rightarrow 0, $$ as $n\rightarrow \infty$. The last inequality holds because $E\bar{X}(1-\bar{X})\leq E\bar{X}$. This implies $\frac{\bar{X}(1-\bar{X})}{\sqrt{n}-1/\sqrt{n}}\overset{P}{\rightarrow }0$.