Hypergeometric function problem

[matematikuvol]

Homework Statement

Calculate
[tex]_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x)[/tex]

Homework Equations

[tex]_2F_1(a,b,c;x)=\sum^{\infty}_{n=0}\frac{(a)_n(b)_n}{n!(c)_n}x^n[/tex]
[tex](a)_n=a(a+1)...(a+n-1)[/tex]

The Attempt at a Solution

[tex](\frac{1}{2})_n=\frac{1}{2}\frac{3}{2}\frac{5}{2}...\frac{2n-1}{2}[/tex]
[tex](\frac{3}{2})_n=\frac{3}{2}\frac{5}{2}\frac{7}{2}...\frac{2n+1}{2}[/tex]
From this relations
[tex]\frac{(\frac{1}{2})_n}{(\frac{3}{2})_n}=\frac{1}{2n+1}[/tex]
But I don't see how to calculate this to the end...

 

[haruspex]

Can you write [tex](\frac{1}{2})_n[/tex] in terms of factorials and powers of 2?

 

[matematikuvol]

Yes I could
[tex](\frac{1}{2})_n=\frac{(2n-1)!}{2^n}[/tex]
[tex](\frac{3}{2})_n=\frac{(2n+1)!}{2^{n-1}}[/tex]
So I get
[tex]_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x)=\sum^{\infty}_{n=0}\frac{(2n-1)!}{(2n+1)2^{n+1}n!}x^n=\sum^{\infty}_{n=0}\frac{(2n-1)!}{2(2n+1)(2n)!}x^n[/tex]
And I don't know what to do know.

 

[lurflurf]

Hint:

[tex]\arcsin (x) \sim \sum_{k=0}^\infty \dfrac{(2k)!x^{2k+1}}{4^k (k!)^2 (2k+1)} \sim \sum_{k=0}^\infty \dfrac{(2k-1)!x^{2k+1}}{(2k)!(2k+1)}[/tex]

I am not sure how to motivate this though. It can be difficult to recognize familiar functions by their power series.

 

[haruspex]

Try writing out the general term of the expansion of (1-x)^-1/2