TITLE: Rate of convergence in the central limit theorem (Lindeberg–Lévy)
QUESTION [19 upvotes]: There are similar posts to this one on stackexchange but none of those seem to actually answer my questions. So consider the CLT in the most common form.
Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of i.i.d. random variables with $X_1 \in L^2(P)$ and $\mathbb{E}[X_i]= \mu$ and $\mathbb{V}ar[X_i] = \sigma^2>0$. Denote with $\widehat{X}:= \frac{(X_1+\dots+X_n)}{n}$. Then it holds that 
$$\sqrt{n} (\widehat{X} - \mu) \overset{\mathscr{D}}{\longrightarrow} \mathscr{N} (0,{\sigma}^2)$$ 
or, equivalently, 
$$\sqrt{n} \left( \frac{\widehat{X} - \mu}{\sigma} \right) \overset{\mathscr{D}}{\longrightarrow} \mathscr{N} (0,1).$$
I often see statements like the rate of convergence is of order $\frac{1}{\sqrt{n}}$. Trying to interpret this, this is what I have understood (informally) so far:

According to the strong law of large numbers, given the above
  conditions, $$\widehat{X} - \mu\overset{a.s.}{\longrightarrow} 0.$$
  However $\widehat{X} - \mu$ stops converging to zero when multiplied
  by $\sqrt{n}$. So one says that the rate of convergence is of order
  $\frac{1}{\sqrt{n}}$.

So here are my questions:

How does one define the order of convergence in this case using formal notation? Why does one say of order $\frac{1}{\sqrt{n}}$ and not of order $\sqrt{n}$?
How do we know that if multiplied, for example, by a factor of lower or higher order, like $\sqrt[3]{n}$ or $n$, one would not get a random variable converging in distribution to some a.s. non-zero random variable (as opposed to the argument: "However $\widehat{X} - \mu$ stops converging to zero when multiplied by $\sqrt{n}$."?
And, most importantly, can someone actually show rigorously that the rate of convergence is exactly of order $\frac{1}{\sqrt{n}}$?

REPLY [4 votes]: This is just a remark that was too long to fit as a comment. The remark is about what people mean when they casually say "$\sqrt{n}$ (or $1/\sqrt{n}$) convergence".
Take $\mu = 0, \sigma = 1$ for simplicity. If $$\frac{1}{\sqrt{n}}\sum_i X_i $$ is "approximately normally distributed", as is guaranteed for sufficiently large $n$ by CLT, then we can approximate deviations from the empirical mean $\frac{1}{n}\sum_i X_i\approx 0$ by using the CLT approximation
$$\mathbb{P}\left(-\frac{\epsilon}{\sqrt{n}} < \frac{1}{n}\sum_i X_i  < \frac{\epsilon}{\sqrt{n}}\right) \approx \mathbb{P}\left(-\epsilon  < N(0,1) < \epsilon\right).\tag{1}$$
Then, heuristically, if you want an extra decimal point of accuracy (i.e. divide $\epsilon$ by 10) with a fixed probability, you need $10^2$ more samples (this keeps the left-hand side of the above approximation asymptotically constant). This is often what people mean when they say "CLT implies $\sqrt{n}$ convergence". 
The slight of hand above is: how large does $n$ have to be? In other words, what is the order of convergence of 
$$\left|\mathbb{P}\left(-\frac{\epsilon}{\sqrt{n}} < \frac{1}{n}\sum_i X_i < \frac{\epsilon}{\sqrt{n}}\right) - \mathbb{P}\left(-\epsilon < N(0,1) < \epsilon\right)\right|.$$
More explicitly, what is the approximation for any set, say
$$\left|\mathbb{P}\left(\frac{1}{\sqrt{n}}\sum_i X_i < x\right) - \mathbb{P}\left(N(0,1) < x\right)\right|?$$ 
It turns out a bound on the above is also of order $1/\sqrt{n}$, and is the content of the Barry-Esseen Theorem, as pointed out in one of the links in the comments.