Given-I-n-0-1-1-x-n-n-e-x-dx-n-N-a-Show-that-x-0-1-1-x-n-e-x-e-and-deduce-that-the-Sequence-I-n-n-converges-to-zero-b-Establish-a-recurrence-relation-between-I-n-and-I-n-

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Question Number 105632 by Ar Brandon last updated on 30/Jul/20

$$\mathrm{Given}{\mathrm{I}}_{\mathrm{n}}={\int }_0^1\frac{{(1-\mathrm{x})}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx},\mathrm{n}\in \mathbb{N}$$$$\mathrm{a}\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{\forall }\mathrm{x}\in [0,1],{(1-\mathrm{x})}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{x}}⩽\mathrm{e}\mathrm{and}\mathrm{deduce}\mathrm{that}\mathrm{the}$$$$\mathrm{Sequence}{\left({\mathrm{I}}_{\mathrm{n}}\right)}_{\mathrm{n}}\mathrm{converges}\mathrm{to}\mathrm{zero}.$$$$\mathrm{b}\mathrm{\setminus }\mathrm{Establish}\mathrm{a}\mathrm{recurrence}\mathrm{relation}\mathrm{between}{\mathrm{I}}_{\mathrm{n}}\mathrm{and}{\mathrm{I}}_{\mathrm{n}+1}$$$$\mathrm{c}\mathrm{\setminus }\mathrm{Deduce}\mathrm{that}\mathrm{e}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\frac{1}{\mathrm{k}!}\right)$$

Answered by mathmax by abdo last updated on 31/Jul/20

$$1)\mathrm{x}\in [0,1]\Rightarrow {\mathrm{e}}^{\mathrm{x}}⩽\mathrm{e}\mathrm{and}{(1-\mathrm{x})}^{\mathrm{n}}⩽1\Rightarrow {(1-\mathrm{x})}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{x}}⩽\mathrm{e}\Rightarrow$$$$\frac{{(1-\mathrm{x})}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{e}}^{\mathrm{x}}⩽\frac{\mathrm{e}}{\mathrm{n}!}\Rightarrow {\int }_0^1\frac{{(1-\mathrm{x})}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}⩽\frac{\mathrm{e}}{\mathrm{n}!}\to 0(\mathrm{n}\to +\mathrm{\infty })\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{I}}_{\mathrm{n}}=0$$$$2)\mathrm{we}\mathrm{have}{\mathrm{I}}_{\mathrm{n}+1}=\frac{1}{(\mathrm{n}+1)!}{\int }_0^1{(1-\mathrm{x})}^{\mathrm{n}+1}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$$$$\Rightarrow (\mathrm{n}+1)!{\mathrm{I}}_{\mathrm{n}+1}={\int }_0^1{(1-\mathrm{x})}^{\mathrm{n}+1}{\mathrm{e}}^{\mathrm{x}}\mathrm{by}\mathrm{parts}\mathrm{u}={(1-\mathrm{x})}^{\mathrm{n}+1}[\mathrm{and}{\mathrm{v}}^{{}^{\prime }}={\mathrm{e}}^{\mathrm{x}}$$$$\Rightarrow {\int }_0^1{(1-\mathrm{x})}^{\mathrm{n}+1}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}={\left[{(1-\mathrm{x})}^{\mathrm{n}+1}{\mathrm{e}}^{\mathrm{x}}\right]}_0^1-{\int }_0^1-(\mathrm{n}+1){(1-\mathrm{x})}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$$$$=-1+(\mathrm{n}+1){\int }_0^1{(1-\mathrm{x})}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=-1+(\mathrm{n}+1)\mathrm{n}!{\mathrm{I}}_{\mathrm{n}}=-1+(\mathrm{n}+1)!{\mathrm{I}}_{\mathrm{n}}$$$$(\mathrm{n}+1)!{\mathrm{I}}_{\mathrm{n}+1}=-1+(\mathrm{n}+1)!{\mathrm{I}}_{\mathrm{n}}\Rightarrow {\mathrm{I}}_{\mathrm{n}+1}={\mathrm{I}}_{\mathrm{n}}-\frac{1}{(\mathrm{n}+1)!}\Rightarrow$$$$\frac{1}{(\mathrm{n}+1)!}={\mathrm{I}}_{\mathrm{n}}-{\mathrm{I}}_{\mathrm{n}+1}\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{(\mathrm{k}+1)!}=\sum _{\mathrm{k}=0}^{\mathrm{n}}({\mathrm{I}}_{\mathrm{k}}-{\mathrm{I}}_{\mathrm{k}+1})\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{1}{\mathrm{k}!}={\mathrm{I}}_{\mathrm{o}}-{\mathrm{I}}_{\mathrm{n}+1}+1\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{1}{\mathrm{k}!}={\mathrm{I}}_0+1\Rightarrow$$$$\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{k}!}={\mathrm{I}}_{\mathrm{o}}+1\mathrm{but}{\mathrm{I}}_0={\int }_0^1{\mathrm{e}}^{\mathrm{x}}=\mathrm{e}-1\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{k}!}=\mathrm{e}$$

Commented by Ar Brandon last updated on 31/Jul/20

Merci monsieur��

Commented by mathmax by abdo last updated on 01/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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