I am trying to prove weak consistency of standard deviation $S$, in other words, $P(|S-\sigma|<\epsilon) \to 1$ as $n \to \infty$ $\forall \epsilon$.
(I have seen in another post that $S$ is a biased estimator of $\sigma$). My attemp is:
Take $\epsilon>0$ then
$P(|S-\sigma|<\epsilon)=P(-\epsilon<S-\sigma<\epsilon)=P((1-\frac{\epsilon}{\sigma})^2<\frac{S^2}{\sigma^2}<(1+\frac{\epsilon}{\sigma})^2)=P((1-\frac{\epsilon}{\sigma})^2(n-1)<\chi^2_{n-1}<(1+\frac{\epsilon}{\sigma})^2(n-1))=$
Using convergence in distribution $\dfrac{\chi^2_n - n}{\sqrt{2n} }\xrightarrow{d}\mathcal N(0,1)$ we get
$=P\bigg{(}\frac{(1-\frac{\epsilon}{\sigma})^2(n-1)-(n-1)}{\sqrt{2(n-1)}}<Z<\frac{(1+\frac{\epsilon}{\sigma})^2(n-1)-(n-1)}{\sqrt{2(n-1)}}\bigg{)}=P\bigg{(}\frac{\sqrt{n-1}((1-\frac{\epsilon}{\sigma})^2-1)}{\sqrt{2}}<Z<\frac{\sqrt{n-1}((1+\frac{\epsilon}{\sigma})^2-1)}{\sqrt{2}}\bigg{)}= P\bigg{(}\frac{\sqrt{n-1}(\frac{-2\epsilon}{\sigma}+(\frac{\epsilon}{\sigma})^2)}{\sqrt{2}}<Z<\frac{\sqrt{n-1}(\frac{2\epsilon}{\sigma}+(\frac{\epsilon}{\sigma})^2)}{\sqrt{2}}\bigg{)}$
But when when $n\to \infty$ the last probability doesn't seem to converge to 1.
I will like to know if I make a mistake in my attemp, or if it is another approach to solve it, because I am beginning to believe that $S$ might not be weakly consistent.
Thank you for your time and help :)!