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

Σ_(n=0) ^∞ (((−1)^n )/(n!(2n+3)))((4/π))^n =?

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

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

e^x =Σ_(n=0) ^∞ (x^n /(n!)) e^(−x^2 ) =Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!)) x^2 e^(−x^2 ) =Σ_(n=0) ^∞ (((−1)^n x^(2n+2) )/(n!)) ∫_0 ^x x^2 e^(−x^2 ) dx=Σ_(n=0) ^∞ (((−1)^n )/(n!))∫_0 ^x x^(2n+2) dx (((√π)erf(x)−2xe^(−x^2 ) )/4)=Σ_(n=0) ^∞ (((−1)^n x^(2n+3) )/(n!(2n+3))) (((√π)erf(x)−2xe^(−x^2 ) )/(4x^3 ))=Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!(2n+3))) let x=(2/( (√π))) Σ_(n=0) ^∞ (((−1)^n )/(n!(2n+3)))((4/π))^n =(((√π)erf((2/( (√π))))−(4/( (√π)))e^(−(4/π)) )/(4×(4/π)×(2/( (√π))))) =((π^2 erf((2/( (√π))))−4πe^(−(4/π)) )/(32))

$${e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!} \\ $$$${e}^{−{x}^{\mathrm{2}} } =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!} \\ $$$${x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{2}} }{{n}!} \\ $$$$\int_{\mathrm{0}} ^{{x}} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{{x}} {x}^{\mathrm{2}{n}+\mathrm{2}} {dx} \\ $$$$\frac{\sqrt{\pi}{erf}\left({x}\right)−\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } }{\mathrm{4}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{3}} }{{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\frac{\sqrt{\pi}{erf}\left({x}\right)−\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } }{\mathrm{4}{x}^{\mathrm{3}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${let}\:{x}=\frac{\mathrm{2}}{\:\sqrt{\pi}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)}\left(\frac{\mathrm{4}}{\pi}\right)^{{n}} =\frac{\sqrt{\pi}{erf}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\right)−\frac{\mathrm{4}}{\:\sqrt{\pi}}{e}^{−\frac{\mathrm{4}}{\pi}} }{\mathrm{4}×\frac{\mathrm{4}}{\pi}×\frac{\mathrm{2}}{\:\sqrt{\pi}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{2}} {erf}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\right)−\mathrm{4}\pi{e}^{−\frac{\mathrm{4}}{\pi}} }{\mathrm{32}} \\ $$

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