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Question Number 128256 by Dwaipayan Shikari last updated on 05/Jan/21

$$1+\frac{1}{2!1!}{\left(\frac{1}{2}\right)}^2+\frac{1}{3!2!}{\left(\frac{1.3}{2^2}\right)}^2+\frac{1}{4!3!}{\left(\frac{1.3.5}{2^3}\right)}^2+....=\frac{4}{\pi }$$$$ProvetheaboveRelation$$

Commented by Dwaipayan Shikari last updated on 06/Jan/21

$$1+\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n^2}{{\left(2\right)}_{n}n!}={}_2{F}_1(\frac{1}{2},\frac{1}{2};2;1)=\frac{\mathrm{\Gamma }\left(1\right)\mathrm{\Gamma }(2-1)}{{\mathrm{\Gamma }}^2\left(\frac{3}{2}\right)}=\frac{4}{\pi }$$

Commented by A8;15: last updated on 06/Jan/21

$$\mathrm{Mr}.\mathrm{can}\mathrm{we}\mathrm{prove}\mathrm{this}\mathrm{relation}\mathrm{with}$$$$\mathrm{continued}\mathrm{fraction}?$$

Commented by Dwaipayan Shikari last updated on 06/Jan/21

$$Heysir!Iamastudent.Kindly{don}^{\prime }tcallmeSir.$$$$IalsothinkthiscanbeprovedbyContinuedfractions$$

Commented by Dwaipayan Shikari last updated on 06/Jan/21

$${tan}^{-1}x=\frac{x}{1+\frac{x^2}{3-x^2+\frac{9x^2}{5-3x^2+\frac{25x^2}{7-5x^2+\frac{49x^2}{..}}}}}$$$$\frac{\pi }{4}=\frac{1}{1+\frac{1}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+...}}}}}\Rightarrow \frac{4}{\pi }=1+\frac{1}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+\frac{9^2}{2+...}}}}}$$

Commented by A8;15: last updated on 06/Jan/21

$$\mathrm{I}\mathrm{respect}\mathrm{you}.\mathrm{May}\mathrm{I}\mathrm{call}\mathrm{you}\mathrm{brother}?$$$$(\mathrm{I}\mathrm{think}\mathrm{I}\mathrm{am}\mathrm{smaller}\mathrm{than}\mathrm{you},\mathrm{so}\mathrm{you}\mathrm{are}\mathrm{my}\mathrm{elder}\mathrm{brother})$$

Commented by Dwaipayan Shikari last updated on 23/Jan/21

$$Myage{e}^{\frac{34}{12}}$$

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