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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.

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2 Answers 2

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$$ {}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\inf}{\frac{(1)_{n}(1)_{n}}{(2)_{n}}\frac{1}{n!}(-x)^{n}}$$

rather than your version:

$$ {}_2F_{1}(1,1;2;-x)=\sum_{n=0}^{\inf}{\frac{(1)_{n}(1)_{n}}{(2)_{n}}(-x)^{n}}$$

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  • $\begingroup$ Thank you, I get it now, I missed that factorial $\endgroup$
    – Hisoka
    May 22, 2021 at 13:25
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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, $$ {_2F_1}(1,1;2;-x)=\sum_{n=0}^\infty\frac{(-x)^n}{n+1}=-\frac{1}{x}\sum_{n=1}^\infty\frac{(-x)^n}{n}=\frac{1}{x}\log(1+x). $$

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