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Question Number 30764 by abdo imad last updated on 25/Feb/18

$$let{I}_n={\int }_0^{\frac{1}{2}}{(1-2t)}^n{e}^{-t}dtwithnintegrnot0$$$$1)provethat\mathrm{\forall }t\in [0,\frac{1}{2}]$$$$\frac{1}{\sqrt{e}}{(1-2t)}^n⩽{(1-2t)}^n{e}^{-t}⩽{(1-2t)}^{n}thenfind{lim}_{n\to \mathrm{\infty }}{I}_n$$$$2)provethat{I}_{n+1}=1-2(n+1)I_n$$$$3)calculate{I}_1,I_2,and{I}_3.$$

Commented by abdo imad last updated on 01/Mar/18

$$1)for0⩽t⩽\frac{1}{2}wehave{(1-2t)}^n⩾01⩽e^t⩽\sqrt{e}\Rightarrow \frac{1}{\sqrt{e}}⩽e^{-t}⩽1$$$$\Rightarrow \frac{1}{\sqrt{e}}{(1-2t)}^n⩽{(1-2t)}^n{e}^{-t}⩽{(1-2t)}^n\Rightarrow$$$$\frac{1}{\sqrt{e}}{\int }_0^{\frac{1}{2}}{(1-2t)}^{n}dt⩽{\int }_0^{\frac{1}{2}}{(1-2t)}^n{e}^{-t}dt⩽{\int }_0^{\frac{1}{2}}{(1-2t)}^{n}dtbut$$$$\frac{1}{\sqrt{e}}{\int }_0^{\frac{1}{2}}{(1-2t)}^{n}dt=\frac{1}{\sqrt{e}}{\left[\frac{-1}{2(n+1)}{(1-2t)}^{n+1}\right]}_0^{\frac{1}{2}}=\frac{1}{2(n+1)\sqrt{e}}$$$$\Rightarrow \frac{1}{2(n+1)\sqrt{e}}⩽I_n⩽\frac{1}{2(n+1)}anditsclearthat$$$${lim}_{n\to \mathrm{\infty }}{I}_n=0$$$$2)letintegratebyparts{u}^{{}^{\prime }}={(1-2t)}^{n}andv=e^{-t}$$$$I_n={\left[\frac{-1}{2(n+1)}{(1-2t)}^{n+1}{e}^{-t}\right]}_0^{\frac{1}{2}}+{\int }_0^{\frac{1}{2}}\frac{-1}{2(n+1)}{(1-2t)}^{n+1}{e}^{-t}dt$$$$=\frac{1}{2(n+1)}-\frac{1}{2(n+1)}{I}_{n+1}\Rightarrow 2(n+1)I_n=1-I_{n+1}\Rightarrow$$$$I_{n+1}=1-2(n+1)I_n$$$$3)I_1=1-2I_{0}but{I}_0={\int }_0^{\frac{1}{2}}{e}^{-t}dt={[-e^{-t}]}_0^{\frac{1}{2}}=1-e^{\frac{-1}{2}}=1-\frac{1}{\sqrt{e}}$$$$I_2=1-4I_1=1-4(1-\frac{1}{\sqrt{e}})=-3+\frac{4}{\sqrt{e}}.$$$$I_3=1-6I_2=1-6(-3+\frac{4}{\sqrt{e}})=19-\frac{24}{\sqrt{e}}.$$$$$$

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