# how to prove the hypergeometric function ${}_2F_{1}(1,1;2;-x)=\frac{\log(1+x)}{x}$

how do I prove that $$\frac{\log(1+x)}{x}={}_2F_{1}(1,1;2;-x)$$ Here is what I tried $${}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\infty}{\frac{(1)_{n}(1)_{n}}{(2)_{n}}(-x)^{n}}$$ next$$(1)_{n}=n!\ \ ,\ \ \ (2)_{n}=(n+1)!$$ $${}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\infty}{\frac{n!n!}{(n+1)!}(-x)^{n}}$$ $${}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\infty}{\frac{n!}{n+1}(-x)^{n}}$$ but $$\frac{\log(1+x)}{x}=\sum_{n=0}^{\infty}{\frac{(-1)^{n}}{n+1}x^{n}}$$ Help me what did I do wrong.

$${}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\inf}{\frac{(1)_{n}(1)_{n}}{(2)_{n}}\frac{1}{n!}(-x)^{n}}$$
$${}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\inf}{\frac{(1)_{n}(1)_{n}}{(2)_{n}}(-x)^{n}}$$
By the definition of the hypergeometric series $${_2F_1}(1,1;2;-x)=\sum_{n=0}^\infty\frac{(1)_n(1)_n}{(2)_n}\frac{(-x)^n}{n!}=\sum_{n=0}^\infty\frac{(1)_n}{(2)_n}(-x)^n,$$ since $$(1)_n=n!$$. Then by the definition of the Pochhammer symbol: $$\frac{(1)_n}{(2)_n}=n!\frac{\Gamma(2)}{\Gamma(n+2)}=\frac{n!}{(n+1)!}=\frac{1}{n+1};$$ whence, $${_2F_1}(1,1;2;-x)=\sum_{n=0}^\infty\frac{(-x)^n}{n+1}=-\frac{1}{x}\sum_{n=1}^\infty\frac{(-x)^n}{n}=\frac{1}{x}\log(1+x).$$