Where does $\sqrt{n}$ come from in central limit theorem (CLT)? A very simple version of central limited theorem as below
$$
 \sqrt{n}\bigg(\bigg(\frac{1}{n}\sum_{i=1}^n X_i\bigg) - \mu\bigg)\ \xrightarrow{d}\ \mathcal{N}(0,\;\sigma^2)
$$
which is Lindeberg–Lévy CLT. I do not understand why there is a $\sqrt{n}$ on the left handside. And Lyapunov CLT says 
$$
\frac{1}{s_n} \sum_{i=1}^{n} (X_i - \mu_i) \ \xrightarrow{d}\ \mathcal{N}(0,\;1)
$$
but why not $\sqrt{s_n}$? Would anyone tell me what are these factors, such $\sqrt{n}$ and $\frac{1}{s_n}$? how do we get them in the theorem?
 A: There is a nice theory of what kind of distributions can be limiting distributions of sums of random variables. The nice resource is the following book by Petrov, which I personally enjoyed immensely.
It turns out, that if you are investigating limits of this type
$$\frac{1}{a_n}\sum_{i=1}^n(X_i-b_i), \quad (1)$$ where $X_i$ are independent random variables, the distributions of limits are only certain distributions.
There is a lot of mathematics going around then, which boils to several theorems which completely characterizes what happens in the limit. One of such theorems is due to Feller:
Theorem Let $\{X_n;n=1,2,...\}$ be a sequence of independent random variables, $V_n(x)$ be the distribution function of $X_n$, and  $a_n$ be a sequence of positive constant. In order that
$$\max_{1\le k\le n}P(|X_k|\ge\varepsilon a_n)\to 0, \text{ for every fixed } \varepsilon>0$$
and
$$\sup_x\left|P\left(a_n^{-1}\sum_{k=1}^nX_k<x\right)-\Phi(x)\right|\to 0$$
it is necessary and sufficient that
$$\sum_{k=1}^n\int_{|x|\ge \varepsilon a_n}dV_k(x)\to 0 \text{ for every fixed }\varepsilon>0,$$
$$a_n^{-2}\sum_{k=1}^n\left(\int_{|x|<a_n}x^2dV_k(x)-\left(\int_{|x|<a_n}xdV_k(x)\right)^2\right)\to 1$$
and
$$a_n^{-1}\sum_{k=1}^n\int_{|x|<a_n}xdV_k(x)\to 0.$$
This theorem then gives you an idea of what $a_n$ should look like.
The general theory in the book is constructed in such way that norming constant is restricted in any way, but final theorems which give necessary and sufficient conditions, do not leave any room for norming constant other than $\sqrt{n}$.
A: Nice question (+1)!!
You will remember that for independent random variables $X$ and $Y$, $Var(X+Y) = Var(X) + Var(Y)$ and $Var(a\cdot X) = a^2 \cdot Var(X)$. So the variance of $\sum_{i=1}^n X_i$ is $\sum_{i=1}^n \sigma^2 = n\sigma^2$, and the variance of $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ is $n\sigma^2 / n^2 = \sigma^2/n$.
This is for the variance. To standardize a random variable, you divide it by its standard deviation. As you know, the expected value of $\bar{X}$ is $\mu$, so the variable
$$ \frac{\bar{X} - E\left( \bar{X} \right)}{\sqrt{ Var(\bar{X}) }} = \sqrt{n} \frac{\bar{X} - \mu}{\sigma}$$ has expected value 0 and variance 1. So if it tends to a Gaussian, it has to be the standard Gaussian $\mathcal{N}(0,\;1)$. Your formulation in the first equation is equivalent. By multiplying the left hand side by $\sigma$ you set the variance to $\sigma^2$.
Regarding your second point, I believe that the equation shown above illustrates that you have to divide by $\sigma$ and not $\sqrt{\sigma}$ to standardize the equation, explaining why you use $s_n$ (the estimator of $\sigma)$ and not $\sqrt{s_n}$.
Addition: @whuber suggests to discuss the why of the scaling by $\sqrt{n}$. He does it there, but because the answer is very long I will try to capture the essense of his argument (which is a reconstruction of de Moivre's thoughts).
If you add a large number $n$ of +1's and -1's, you can approximate the probability that the sum will be $j$ by elementary counting. The log of this probability is proportional to $-j^2/n$. So if we want the probability above to converge to a constant as $n$ goes large, we have to use a normalizing factor in $O(\sqrt{n})$.
Using modern (post de Moivre) mathematical tools, you can see the approximation mentioned above by noticing that the sought probability is
$$P(j) = \frac{{n \choose n/2+j}}{2^n} = \frac{n!}{2^n(n/2+j)!(n/2-j)!}$$
which we approximate by Stirling's formula
$$ P(j) \approx \frac{n^n e^{n/2+j} e^{n/2-j}}{2^n e^n (n/2+j)^{n/2+j} (n/2-j)^{n/2-j} } = \left(\frac{1}{1+2j/n}\right)^{n+j} \left(\frac{1}{1-2j/n}\right)^{n-j}. $$
$$ \log(P(j)) = -(n+j) \log(1+2j/n) - (n-j) \log(1-2j/n) \\
\sim -2j(n+j)/n + 2j(n-j)/n \propto -j^2/n.$$
A: $s_n$ represents the sample standard deviation for the sample mean. $s_n$$^2$ is the sample variance for the sample mean and it equals $S_n^2/n$, where $S_n^2$ is the sample estimate of the population variance. Since $s_n =S_n/\sqrt{n}$ that explains how $\sqrt n$ appears in the first formula.  Note there would be a $\sigma$ in the denominator if the limit were $N(0,1)$ but the limit is given as $N(0, \sigma^2)$.  Since $S_n$ is a consistent estimate of $\sigma$ it is used in the second equation to take $\sigma$ out of the limit.
A: Intuitively, if $Z_n \to \mathcal N(0, \sigma^2)$ for some $\sigma^2$ we should expect that $\mbox{Var}(Z_n)$ is roughly equal to $\sigma^2$; it seems like a pretty reasonable expectation, though I don't think it is necessary in general. The reason for the $\sqrt n$ in the first expression is that the variance of $\bar X_n - \mu$ goes to $0$ like $\frac 1 n$ and so the $\sqrt n$ is inflating the variance so that the expression just has variance equal to $\sigma^2$. In the second expression, the term $s_n$ is defined to be $\sqrt{\sum_{i = 1} ^ n \mbox{Var}(X_i)}$ while the variance of the numerator grows like $\sum_{i = 1} ^ n \mbox{Var}(X_i)$, so we again have that the variance of the whole expression is a constant ($1$ in this case).
Essentially, we know something "interesting" is happening with the distribution of $\bar X_n := \frac 1 n \sum_i X_i$, but if we don't properly center and scale it we won't be able to see it. I've heard this described sometimes as needing to adjust the microscope. If we don't blow up (e.g.) $\bar X - \mu$ by $\sqrt n$ then we just have $\bar X_n - \mu \to 0$ in distribution by the weak law; an interesting result in it's own right but not as informative as the CLT. If we inflate by any factor $a_n$ which is dominated by $\sqrt n$, we still get $a_n(\bar X_n - \mu) \to 0$ while any factor $a_n$ which dominates $\sqrt n$ gives $a_n(\bar X_n - \mu) \to \infty$. It turns out $\sqrt n$ is just the right magnification to be able to see what is going on in this case (note: all convergence here is in distribution; there is another level of magnification which is interesting for almost sure convergence, which gives rise to the law of iterated logarithm).
