prove $\lim \limits_{n \to \infty}{\frac{W_n\sqrt{n}}{(n-1)\sqrt{2}}} \sim N_{(0,1)}$ I searched the internet alot . The only relevant clue is in Wikipedia:
F-distribution
Beside that I didn't find any proof for this theory.   

If $Y$ has $B\left(\frac{d_1}{2}, \frac{d_2}{2}\right)$ distribution, then show that $X$ with given formula has $F\left(d_1,d_2\right)$ distribution.
$$X = \frac{d_2Y}{d_1\left(1-Y\right)}$$
I tried this but I can't do anything because of $\Gamma{}$ integral:
$$Y = \frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}$$ 
so:
$$X = \frac{d_2Y}{d_1\left(1-Y\right)}$$
$$X = \frac{d_2\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}}{d_1\left(1-\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}\right)}$$
$$X = x^{\frac{d_1}{2} - 1}\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}\left(\frac{d_2\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)\left(1-x\right)^{\frac{d_2}{2}-1}}{d_1\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right) - \Gamma\left({\frac{d_1 + d_2}{2}}\right)x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}}\right)$$ 
compare it to $F\left(d_1,d_2\right)$:  
$$F\left(d_1,d_2\right)=x^{\frac{d_1}{2} - 1}\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}\left(\frac{d_1}{d_2}\right)^{\frac{d_1}{2}-1}\frac{1}{\left(1+\frac{d_1}{d_2}x\right)^{\frac{d_1+d_2}{2}}}$$ 
so:
$$\left(\frac{d_1}{d_2}\right)^{\frac{d_1}{2}-1}\frac{1}{\left(1+\frac{d_1}{d_2}x\right)^{\frac{d_1+d_2}{2}}} = \left(\frac{d_2\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)\left(1-x\right)^{\frac{d_2}{2}-1}}{d_1\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right) - \Gamma\left({\frac{d_1 + d_2}{2}}\right)x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}}\right)$$
I can't go on anymore because of $\Gamma$ integral!  
If possible, give me a hint to prove this.
Any help will be appreciated.
 A: I ask my teacher and it was wrong. The question is:  
Problem
Show that:
If $X_1,X_2,...,X_n$ be independent random variable with $\chi^2_1$ (df=1) and $W_n=X_1+X_2+...+X_n$ then
$$\lim \limits_{n \to \infty}{\frac{(W_n - n)\sqrt{n}}{n\sqrt{2}}} \sim N_{(0,1)}$$
Answer
based on my last proof in question:
I start with this fact that $\chi^2_1$ has $\mu=1$ and $\sigma^2=2$ and based on CLT (central limit theorem), when $n \to \infty$ we have: 
$$W_n=X_1+X_2+...+X_n \sim N_{(n\mu, n\sigma^2)} \sim N_{(n,2n)}$$
(based on my teacher lesson, we write $\sigma^2$ in second paramter of $N$).  
So we know that:
$$Z=\frac{W_n-n}{\sqrt{2n}} \sim N_{(0,1)}$$
and we can write it:
$$W_n=\sqrt{2n}N_{(0,1)}+n$$
Now, we go to solve the limit:
$$\lim \limits_{n \to \infty}{\frac{(W_n - n)\sqrt{n}}{n\sqrt{2}}} = 
\lim \limits_{n \to \infty}{\frac{(\sqrt{2n}N_{(0,1)})\sqrt{n}}{n\sqrt{2}}} = 
N_{(0,1)}\lim \limits_{n \to \infty}{\frac{(\sqrt{2n})\sqrt{n}}{n\sqrt{2}}} = 
N_{(0,1)}$$
proof finished!
I'm sorry but it is a mistake from my teacher. However thanks for help.  
