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Question Number 56060 by necx1 last updated on 09/Mar/19

$$\int e^{-x^2}dxasaninfiniteseries.Hence$$$$investigateitsconverge.$$

Commented by maxmathsup by imad last updated on 09/Mar/19

$$wehave{e}^{-x^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{n!}\Rightarrow \int e^{-x^2}dx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}\int x^{2n}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)n!}{x}^{2n+1}letfindRtheradiusofconvergene$$$$let{u}_n\left(x\right)=\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)n!}\Rightarrow forx\ne 0$$$$\mid \frac{u_{n+1}\left(x\right)}{u_{n\left(x\right)}}\mid =\mid \frac{\mid x{\mid }^{2n+3}}{(2n+3)!(n+1)!}\frac{(2n+1)n!}{x^{2n+1}}\mid =\frac{2n+1}{(2n+3)(n+1)}\mid x{\mid }^2\to 0(n\to +\mathrm{\infty })\Rightarrow$$$$\Rightarrow R=+\mathrm{\infty }.$$

Commented by necx1 last updated on 09/Mar/19

$$ThanksforthehelpthoIreallydonot$$$$understand.Pleaseshedmorelight$$

Commented by maxmathsup by imad last updated on 09/Mar/19

$$sirtakealookatcknvergenceofseriesubject...$$

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