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Question Number 141946 by mnjuly1970 last updated on 27/May/21

$$....nicecalculus...$$$${lim}_{n\to \mathrm{\infty }}n{\int }_0^1{\left(\frac{2x}{1+x}\right)}^n=???$$

Answered by mnjuly1970 last updated on 29/May/21

$$\mathit{\varphi }={lim}_{n\to \mathrm{\infty }}n.2^n{\int }_0^1{\left(\frac{x}{1+x}\right)}^n\stackrel{?}{=}2$$$$solution::t:=\frac{2x}{1+x}\Rightarrow x=\frac{t}{2-t}$$$$\mathit{\varphi }:={lim}_{n\to \mathrm{\infty }}\frac{n}{2}{\int }_0^1{t}^n.\frac{1}{{(1-\frac{t}{2})}^2}dt$$$$:={lim}_{n\to \mathrm{\infty }}\frac{n}{2}{\int }_0^1{t}^n\left(\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{t^k}{2^k}\right)\left(\underset{m=0}{\sum ^{\mathrm{\infty }}}\frac{t^m}{2^m}\right)dt$$$$:={lim}_{n\to \mathrm{\infty }}\frac{n}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}\underset{m=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k.2^m}\left(\frac{1}{n+m+m}\right)$$$$:=\frac{1}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}\left(\frac{1}{1-\frac{1}{2}}\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}=\frac{1}{1-\frac{1}{2}}=2$$$$m.n$$$$$$

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