n-0-1-n-n-2n-3-4-pi-n-

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

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n! (2 n + 3)} {(\frac{4}{π})}^{n} = ?

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

e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

e^{- x^{2}} = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n}}{n!}

x^{2} e^{- x^{2}} = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n + 2}}{n!}

\int_{0}^{x} x^{2} e^{- x^{2}} d x = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int_{0}^{x} x^{2 n + 2} d x

\frac{\sqrt{π} e r f (x) - 2 {x e}^{- x^{2}}}{4} = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n + 3}}{n! (2 n + 3)}

\frac{\sqrt{π} e r f (x) - 2 {x e}^{- x^{2}}}{4 x^{3}} = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n}}{n! (2 n + 3)}

l e t x = \frac{2}{\sqrt{π}}

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n! (2 n + 3)} {(\frac{4}{π})}^{n} = \frac{\sqrt{π} e r f (\frac{2}{\sqrt{π}}) - \frac{4}{\sqrt{π}} e^{- \frac{4}{π}}}{4 \times \frac{4}{π} \times \frac{2}{\sqrt{π}}}

= \frac{π^{2} e r f (\frac{2}{\sqrt{π}}) - 4 π e^{- \frac{4}{π}}}{32}

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