Showing $\sqrt{S^2}$ is a consistent estimator of standard deviation $\sigma$. Given $X_1,\dots,X_n\stackrel{iid}{f}(x;\mu,\sigma^2)$ with population mean $\mu$ and population standard deviation $\sigma$. I want to show $\sqrt{S^2}$ is a consistent estimator for $\sigma$. So if I read my book correctly, we want to show
$$\lim_{n\rightarrow\infty}P(|\sqrt{S_n^2}-\sigma|>\epsilon)=0\quad\forall\epsilon >0 $$
So invoking Chebyshev's inequality we have
$$
\begin{align*}
P((\sqrt{S_n^2}-\sigma)^2>\epsilon^2)\leq&\frac{\mathbb{E}(\sqrt{S_n^2}-\sigma)}{\epsilon^2}
\\
={}&\frac{\mathrm{Var}(\sqrt{S_n^2})}{\epsilon^2}\\
={}&\frac{\mathbb{E}[(\sqrt{S_n^2})^2]-[\mathbb{E}(\sqrt{S_n^2})]^2}{\epsilon^2}\\
={}&\frac{\mathbb{E}(S_n^2)-[\mathbb{E}(\sqrt{S_n^2})]^2}{\epsilon^2}\\
={}&\frac{\sigma^2-[\mathbb{E}(\sqrt{S_n^2})]^2}{\epsilon^2}\text{ (Since $S^2$ is an unbiased estimator)}
\end{align*}
$$
This is where I get stuck because I know nothing about the distribution besides it being $\textit{iid}$, so I have no idea what to do with the $[\mathbb{E}(\sqrt{S^2})]^2$ term. I know I need to have a term with $n$ to take a limit, but I am unsure if I have ever taken the expected value of a sample standard deviation or if I even know how to properly. Any help or guidance would be appreciated. Just to save time Jensen's inequality is something that I have not learned in the course yet.
 A: I have done a little more research - here is my answer.
Given $X_1,\dots,X_n\stackrel{iid}{f}(x;\mu,\sigma^2)$, we know that by the weak law of large numbers:
$$\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2\stackrel{p}{\longrightarrow} \sigma^2$$
By the continuous mapping theorem, the square root function preserves continuity, thus
$$\sqrt{\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2}\stackrel{p}{\longrightarrow} \sqrt{\sigma^2}$$
$$\Rightarrow \sqrt{S^2}\stackrel{p}{\longrightarrow}\sigma$$
It was more theory heavy and that is where I am weak, this has been a good learning experience.
A: Henry said it all, but I will elaborate here. The sample variance is
\begin{equation}
S^2 = \frac{1}{n-1} \sum_{i=1}^n \left(  X_i - \overline{X} \right)^2,
\end{equation}
and since $X_i \sim f(x;\mu,\sigma)$, the population variance is
\begin{equation}
\sigma^2 = \frac{1}{n} \sum_{i=1}^n \left(  X_i - \mu \right)^2.
\end{equation}
It is obvious that $\mathbb{E}(\overline{X}) = \mu$ since $X_i \sim f(x;\mu,\sigma)$,
\begin{equation}
\lim_{n\to\infty} \frac{\sigma^2}{S^2} = \lim_{n\to\infty} \frac{n-1}{n} = 1.
\end{equation}
