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Question Number 146087 by mathmax by abdo last updated on 10/Jul/21

$$1)\mathrm{find}{\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-2\mathrm{x}}\mathrm{dx}$$$$2)\mathrm{nature}\mathrm{of}\mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}?$$

Answered by ArielVyny last updated on 24/Jul/21

$$U_n={\int }_0^1{x}^n{e}^{-2x}dx$$$$x^n{e}^{-2x}=\sum _{n⩾0}\frac{{(-1)}^n{2}^n{x}^{2n}}{n!}$$$${\int }_0^1{x}^n{e}^{-2x}dx=\sum _{n⩾0}\frac{{(-2)}^n}{n!}\left[\frac{1}{2n+1}\right]$$$$U_n=\sum _{n⩾0}\frac{{(-2)}^n}{n!(2n+1)}=\sum _{n⩾0}\frac{{(-1)}^n{2}^n}{(2n+1)n!}$$$$U_{n}CV$$$$$$

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