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Question Number 138598 by qaz last updated on 15/Apr/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!(2n+3)}{\left(\frac{4}{\pi }\right)}^n=?$$

Answered by mr W last updated on 15/Apr/21

$$e^x=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$e^{-x^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{2n}}{n!}$$$$x^2{e}^{-x^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{2n+2}}{n!}$$$${\int }_0^x{x}^2{e}^{-x^2}dx=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}{\int }_0^x{x}^{2n+2}dx$$$$\frac{\sqrt{\pi }erf\left(x\right)-2{xe}^{-x^2}}{4}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{2n+3}}{n!(2n+3)}$$$$\frac{\sqrt{\pi }erf\left(x\right)-2{xe}^{-x^2}}{4x^3}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{2n}}{n!(2n+3)}$$$$letx=\frac{2}{\sqrt{\pi }}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!(2n+3)}{\left(\frac{4}{\pi }\right)}^n=\frac{\sqrt{\pi }erf\left(\frac{2}{\sqrt{\pi }}\right)-\frac{4}{\sqrt{\pi }}{e}^{-\frac{4}{\pi }}}{4\times \frac{4}{\pi }\times \frac{2}{\sqrt{\pi }}}$$$$=\frac{{\pi }^{2}erf\left(\frac{2}{\sqrt{\pi }}\right)-4\pi e^{-\frac{4}{\pi }}}{32}$$

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