We have the 1-d symmetric random walk $X_{n}$ with $X_{0}=0$ such that$$X_{n}=X_{0}+\sum_{i=1}^{n} Z_{n}$$ Where $Z_{n}\in\{-1,1\}$ with $\mathbb{P}(\mathbb Z_{n}=-1)=\mathbb Z_{n}=1=1/2$.
We can calculate the probability $$\mathbb{P}\left(X_{2 n}=2 j\right)=\left(\begin{array}{c} {2 n} \\ {n+j} \end{array}\right) 2^{-2 n}=\frac{(2 n) !}{(n+j) !(n-j) !} 2^{-2 n}$$ By using stirling's formula, when $n\rightarrow \infty$, we have $$\lim _{n \rightarrow \infty} \mathbb{P}\left(X_{2 n}=2 j\right)\approx \frac{1}{\sqrt{n \pi}} e^{-\frac{j^{2}}{n}}$$ this probability is like normal distribution but how to write in this form $$\mathbb{P}(a<X_{2n}<b)=...$$ for some interval $(a,b)$ and how to prove it
The last question is that how to prove that $X_{n}$ is normally distributed. I know the method is by using CLT, but CLT is dealt with $\bar{X}$, which confused me.
$X_n$ is not normally distributed. The sense in which the limiting distribution is normal is exactly that given by the Central Limit Theorem: $X_n/\sqrt{n}$ tends in distribution to a normal distribution.