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Question Number 141946 by mnjuly1970 last updated on 27/May/21
....nice calculus... lim_(nā†’āˆž) nāˆ«_0 ^( 1) (((2x)/(1+x)))^n =???
$$\:\:\:\:\:\:\:\:\:\:….{nice}\:\:\:{calculus}… \\ $$$$\:\:\:\:{lim}_{{n}\rightarrow\infty} {n}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}}\right)^{{n}} =??? \\ $$
Answered by mnjuly1970 last updated on 29/May/21
š›—=lim_(nā†’āˆž) n.2^n āˆ«_0 ^( 1) ((x/(1+x)))^n =^? 2 solution:: t:=((2x)/(1+x)) ā‡’ x=(t/(2āˆ’t)) š›—:=lim_(nā†’āˆž) (n/2)āˆ«_0 ^( 1) t^n .(1/((1āˆ’(t/2))^2 ))dt := lim_(nā†’āˆž) (n/2)āˆ«_0 ^( 1) t^n (Ī£_(k=0) ^āˆž (t^k /2^k ))(Ī£_(m=0) ^āˆž (t^m /2^m ))dt :=lim_(nā†’āˆž) (n/2)Ī£_(k=0) ^āˆž Ī£_(m=0) ^āˆž (1/(2^k .2^m ))((1/(n+m+m))) :=(1/2)Ī£_(k=0) ^āˆž (1/2^k )((1/(1āˆ’(1/2))))=Ī£_(k=0) ^āˆž (1/2^k )=(1/(1āˆ’(1/2)))=2 m.n
$$\:\:\:\:\:\:\:\boldsymbol{\phi}={lim}_{{n}\rightarrow\infty} {n}.\mathrm{2}^{{n}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{x}}{\mathrm{1}+{x}}\right)^{{n}} \overset{?} {=}\mathrm{2} \\ $$$$\:\:\:\:\:\:{solution}::\:{t}:=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}}\:\Rightarrow\:{x}=\frac{{t}}{\mathrm{2}āˆ’{t}} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}:={lim}_{{n}\rightarrow\infty} \frac{{n}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} .\frac{\mathrm{1}}{\left(\mathrm{1}āˆ’\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} }{dt} \\ $$$$\:\:\:\::=\:\:{lim}_{{n}\rightarrow\infty} \frac{{n}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} \left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{{t}^{{k}} }{\mathrm{2}^{{k}} }\right)\left(\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{t}^{{m}} }{\mathrm{2}^{{m}} }\right){dt} \\ $$$$\:\:\::={lim}_{{n}\rightarrow\infty} \frac{{n}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} .\mathrm{2}^{{m}} }\left(\frac{\mathrm{1}}{{n}+{m}+{m}}\right) \\ $$$$\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\left(\frac{\mathrm{1}}{\mathrm{1}āˆ’\frac{\mathrm{1}}{\mathrm{2}}}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }=\frac{\mathrm{1}}{\mathrm{1}āˆ’\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2} \\ $$$$\:\:\:\:{m}.{n} \\ $$$$\:\:\:\:\: \\ $$

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