Generalization of Central Limit Theorem? In my probability class we learned the Central Limit Theorem in the following form.

Theorem: Let $\{X_i\}_{i=1}^\infty$ be a sequence of independent identically distributed random variables and suppose that $E(X_i)=\mu$, $\operatorname{Var}(X_i)=\sigma^2<\infty$. Then,

$$\sqrt{n}\frac{\frac{1}{n}\sum_{i=1}^nX_i-\mu}{\sigma}\stackrel{P}{\longrightarrow}N(0,1),\quad\text{as }n\to\infty$$
in probability, where the notation $N(\mu,\sigma^2)$ means normal distribution of mean $\mu$ and variance $\sigma^2$.
Now, if we remove $\sigma$ and $\mu$, do we get
$$\sqrt{n}\frac{1}{n}\sum_{i=1}^nX_i\stackrel{P}{\longrightarrow}N(\mu,\sigma^2)?$$
This seems very natural, but I cannot prove it. Is it true? I think I can prove it by removeing only $\sigma$, but I am not sure for $\mu$.
 A: $$\sqrt{n}\frac{\frac{1}{n}\sum_{i=1}^nX_i-\mu}{\sigma}\stackrel{d}{\longrightarrow}N(0,1),\quad\text{as }n\to\infty$$ $$\implies \sqrt{n}\left(\frac{1}{n}\sum_{i=1}^nX_i-\mu\right)\stackrel{d}{\longrightarrow}N(0,\sigma^2),\quad\text{as }n\to\infty$$ $$\implies \frac{1}{\sqrt{n}}\sum_{i=1}^nX_i-(\sqrt{n}-1)\mu \stackrel{d}{\longrightarrow}N(\mu,\sigma^2),\quad\text{as }n\to\infty$$ but I am not sure that is more useful
A: Be glad that you have not been able to prove that 
$\displaystyle\sqrt{n}\frac{1}{n}\sum_{i=1}^nX_i\stackrel{P}{\longrightarrow}N(\mu,\sigma^2)$
because the result is not true, not only for the desired mode of convergence
but also for any other mode of convergence. As my comment on your
question pointed out,
$$\displaystyle E\left[\sqrt{n}\frac 1n\sum_{i=1}^nX_i\right] = \sqrt{n}\frac 1n E\left[\sum_{i=1}^nX_i\right] = \sqrt{n}\frac 1n n\mu = \sqrt{n}\mu
\to \infty ~\text{as}~ n \to \infty$$
and so the random variable $\displaystyle\sqrt{n}\frac{1}{n}\sum_{i=1}^nX_i$
does not converge to a fixed random variable in any sense.
