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Question Number 80823 by M±th+et£s last updated on 06/Feb/20

$$showthat$$$$\underset{x\to \mathrm{\infty }}{lim}{H}_n=2F_1(1,1;2,1)$$$$$$$$ln\left(4\right)-2ln\left(3\right)=2F_1(1,1;2;\frac{-1}{2})$$

Answered by ~blr237~ last updated on 07/Feb/20

$$knowingthat{}_2{F}_1(a,b,c,z)=\frac{1}{B(b,c-b)}{\int }_0^1{x}^{b-1}{(1-x)}^{c-b-1}{(1-zx)}^{-a}dx$$$${}_2{F}_1(1,1,2,1)=\frac{1}{B(1,2-1)}{\int }_0^1{x}^{1-1}{(1-x)}^{2-1-1}{(1-1\times x)}^{-1}dx$$$$=\frac{\mathrm{\Gamma }\left(2\right)}{\mathrm{\Gamma }\left(1\right)\mathrm{\Gamma }\left(1\right)}{\int }_0^1\frac{1}{1-x}dx$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^{n}dx=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}=\underset{n\to \mathrm{\infty }}{\lim }\underset{p=1}{\sum ^{n+1}}\frac{1}{p}$$$${}_2{F}_1(1,1,2,-\frac{1}{2})=\frac{1}{B(1,1)}{\int }_0^1\frac{1}{1+\frac{x}{2}}dx={\left[2ln(1+\frac{x}{2})\right]}_0^1=2ln\left(\frac{3}{2}\right)=2ln3-ln4$$

Commented by M±th+et£s last updated on 07/Feb/20

$$thankyousomuchsir.great$$

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