Proving a remainder term converges to 0 in probability So we have these definitions:
σ̂^2_1= (1/n)∑(Xi−μ)^2 
σ̂^2_2= (1/n)∑(Xi−Xbar)^2 
I have shown that 
n^0.5(σ̂^2_2−σ^2)= n^0.5(σ̂^2_1−σ^2)- n^0.5(Xbar-μ)^2
I am trying to show that the remainder term - n^0.5(Xbar-μ)^2 converges in probability to 0.
My idea is:
Xbar -> μ in probability, so Xbar -> μ in Law since they equivalent when converging to a constant.
μ ->μ in probability, so therefore by Slutsky's theorem Xbar -μ -> 0 in law, and therefore again it would converge to 0 in probability since it is to a constant.
Then by the continuous mapping theorem, letting g(x)=x^2, this would mean that (Xbar- μ)^2 ->0 in probability. 
And then finally therefore n^0.5(Xbar-μ) -> 0 in probability. 
Are these statements correct, if not, can you explain why they are wrong and what it should be? Thank you!
 A: 
And then finally $\sqrt{n}(\bar{X} - \mu)^2 \rightarrow 0$ in probability. 

You have to be careful with this step.  As a general rule, the fact that $f(X_1,...,X_n) \rightarrow 0$ does not imply $\sqrt{n} \cdot f_n(X_1,...,X_n) \rightarrow 0$ since the term $\sqrt{n} \rightarrow \infty$ gives you an indeterminate form.  In general, the latter term can cause the convergence to zero to fail, since it gets bigger and bigger as $n \rightarrow \infty$.  You therefore have to establish the limiting result by showing that $f_n(X_1,...,X_n)$ goes to zero "faster" than $\sqrt{n}$ goes to infinity.
In this particular case, proving the desired convergence result requires you to assume that the sample values come from an underlying IID form and that the kurtosis of the underlying distribution is finite.  To prove the result we let $\kappa \equiv \mathbb{E}((X_i-\mu)^4) / \sigma^4$ denote the kurtosis of the underlying sampling distribution.  (In the proof below we will also add $n$ as a subscript to the sample mean, to show this dependence explicitly.)

Theorem: If $X_1,X_2,X_3,... \sim \text{IID Dist}$ with finite kurtosis $\kappa < \infty$ then:
$$\sqrt{n}(\bar{X}_n - \mu)^2 \overset{P}{\rightarrow} 0.$$ 
Proof: Since $\kappa < \infty$ the variance exists (i.e., $\sigma < \infty$) and so we can write the term of interest as:
$$\begin{equation} \begin{aligned}
H(n) \equiv \sqrt{n} \cdot (\bar{X}_n - \mu)^2 
= \frac{\sigma^2}{\sqrt{n}} \cdot \Bigg( \frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}} \Bigg)^2 
= \frac{\sigma^2}{\sqrt{n}} \cdot Z_n^2,
\end{aligned} \end{equation}$$
where $Z_n$ is the standardised sample mean, which has $\mathbb{E}(Z_n) = 0$ and $\mathbb{S}(Z_n) = 1$.  With a bit of algebra it can easily be shown that $\mathbb{E}(Z_n^2) = 1$ and $\mathbb{V}(Z_n^2) = \kappa-1$.  As a preliminary matter, we also note that the triangle inequality gives us $|Z_n^2-0| \leqslant |Z_n^2-1| + 1$.  Hence, for all $\varepsilon > 0$ we can use Chebyshev's inequality to obtain:
$$\begin{equation} \begin{aligned}
\mathbb{P}(|H(n) - 0| < \varepsilon)
&= 1 - \mathbb{P}(|H(n) - 0| \geqslant \varepsilon) \\[6pt]
&= 1 - \mathbb{P} \Bigg( \frac{\sigma^2}{\sqrt{n}} |Z_n^2 - 0| \geqslant \varepsilon \Bigg) \\[6pt]
&= 1 - \mathbb{P} \Bigg( |Z_n^2 - 0| \geqslant \frac{\sqrt{n} \cdot \varepsilon}{\sigma^2} \Bigg) \\[6pt]
&\geqslant 1 - \mathbb{P} \Bigg( |Z_n^2 - 1| +1 \geqslant \frac{\sqrt{n} \cdot \varepsilon}{\sigma^2} \Bigg) \\[6pt]
&= 1 - \mathbb{P} \Bigg( |Z_n^2 - 1| \geqslant \frac{\sqrt{n} \cdot \varepsilon}{\sigma^2} - 1 \Bigg) \\[6pt]
&= 1 - \mathbb{P} \Bigg( |Z_n^2 - 1| \geqslant \frac{\sqrt{n} \cdot \varepsilon - \sigma^2}{\sigma^2} \Bigg) \\[6pt]
&= 1 - \mathbb{P} \Bigg( |Z_n^2 - \mathbb{E}(Z_n^2)| \geqslant \frac{\sqrt{n} \cdot \varepsilon - \sigma^2}{\sigma^2 \sqrt{\kappa-1}} \cdot \mathbb{S}(Z_n^2) \Bigg) \\[6pt]
&= 1 - \Big( \frac{\sigma^2 \sqrt{\kappa-1}}{\sqrt{n} \cdot \varepsilon - \sigma^2} \Big)^2. \\[6pt]
\end{aligned} \end{equation}$$
Now, taking limits (and reminding ourselves that $\sigma < \infty$ and $\kappa < \infty$) we get:
$$\lim_{n \rightarrow \infty} \mathbb{P}(|H(n) - 0| < \varepsilon)
\geqslant 1 - \lim_{n \rightarrow \infty} \Big( \frac{\sigma^2 \sqrt{\kappa-1}}{\sqrt{n} \cdot \varepsilon - \sigma^2} \Big)^2 = 1-0 = 1.$$
This is the explicit requirement for $H(n) \overset{P}{\rightarrow} 0$, so we have established the theorem.  $\blacksquare$

If you would like to know more about moments and asymptotic forms for the standard sampling statistics, you can find a number of results in O'Neill (2014).  The above proof is a standard method for showing convergence in probability (see e.g., Result 14, pp. 285, 293-294).  As can be seen from the above theorem and proof, for the IID model, the requirement of finite kurtosis is sufficient to yield the desired limiting result.  For distributions with infinite kurtosis the result may not hold.
