Convergence in distribution for a maximum likelihood estimator

I need some help with the following exercise. Let $$X_1,\dots,X_n$$ normal random variables i.i.d. with known mean $$\mu$$ and unknown variance $$\sigma^2$$.

1. Find $$\sigma$$ maximum likelihood estimator $$T_n$$;
2. Prove that $$T_n$$ is consistent;
3. Find the asymptotic distribution of $$\sqrt{n}(T_n-\sigma)$$.

Now, 1. and 2. are very easy:

1. some calculation and the invariance principle for ML estimators show us that $$T_n=\sqrt{\frac1n\sum_{k=1}^n(X_i-\mu)^2}$$;
2. for the weak law of large numbers $$T_n^2$$ converges in probability to $$\sigma^2$$, so $$T_n\rightarrow \sigma$$ in probability.

I'm actually stuck with the third request. I've tried applying the Cramer theorem, but I failed in finding a function $$F(x)\in L^1(\mathbb{R})$$ such that $$\forall\sigma>0\quad\left|\frac{d}{d\sigma}f(\sigma;x)\right|\leq F(x),\qquad \text{where }f\text{ is the density function of }X_1.$$ I also tried finding the $$T_n$$ characteristic function $$\phi_{T_n}$$ and computing $$\lim_{n\to+\infty}\phi_{\sqrt{n}(T_n-\sigma)},$$ but I gained nothing. Can someone help me?

Cramer theorem can be found here: Mathematical Methods of Statistics - Harald Cramer, Princeton University Press, 1946, page 500.

• maybe a hint, not sure: $\sqrt{n} T_n / \sigma$ is $\chi$ distributed with $n$ degrees of freedom Commented Nov 2, 2019 at 18:08

Notice that since $$\mu$$ is known, $$\sum_{i=1}^n \frac{(X_i-\mu)^2}{\sigma^2}=\frac{nT_n^2}{\sigma^2}\sim \chi^2_n$$
So by classical CLT, you have $$\sqrt n(T_n^2-\sigma^2)\stackrel{L}\longrightarrow N\left(0,2\sigma^4\right)$$
Now simply apply 'Delta method' to get the asymptotic behaviour of $$\sqrt n(T_n-\sigma)$$.
$$\sqrt n(T_n-\sigma)\stackrel{L}\longrightarrow N\left(0,\frac{1}{I(\sigma)}\right)\,,$$
where $$I(\sigma)=\mathbb E_{\sigma}\left[\frac{\partial}{\partial\sigma}\ln f_{\sigma}(X)\right]^2$$ is the Fisher information in $$X\sim N(\mu,\sigma^2)$$ having pdf $$f_{\sigma}$$.