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Question-145783




Question Number 145783 by SOMEDAVONG last updated on 08/Jul/21
Answered by mathmax by abdo last updated on 08/Jul/21
let U_n ={(1+(1/n))(1+(2/n))...(1+(n/n))^(1/n) ⇒logu_n =(1/n)log(Π_(k=1) ^n  (1+(k/n))) =(1/n)Σ_(k=1) ^n  log(1+(k/n))  ⇒lim_(n→+∞) logU_n =∫_0 ^1  log(1+x)dx =_(1+x=t)   ∫_1 ^2  logt dt  =[tlogt−t]_1 ^2  =2log2−2+1 =2log2−1 ⇒  lim_(n→+∞) U_n =e^(2log2−1)  =(4/e)=L
$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{n}}\right)…\left(\mathrm{1}+\frac{\mathrm{n}}{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \Rightarrow\mathrm{logu}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)\right. \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{logU}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}\:=_{\mathrm{1}+\mathrm{x}=\mathrm{t}} \:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\mathrm{logt}\:\mathrm{dt} \\ $$$$=\left[\mathrm{tlogt}−\mathrm{t}\right]_{\mathrm{1}} ^{\mathrm{2}} \:=\mathrm{2log2}−\mathrm{2}+\mathrm{1}\:=\mathrm{2log2}−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\mathrm{2log2}−\mathrm{1}} \:=\frac{\mathrm{4}}{\mathrm{e}}=\mathrm{L} \\ $$
Answered by puissant last updated on 08/Jul/21
L=lim_(n→+∞) Π_(k=1) ^n (1+(k/n))^(1/n) ;  U=ln(L)  U=lim_(n→+∞) (1/n)Σ_(k=1) ^n ln(1+(k/n))=f((k/n))  =∫_0 ^1 ln(1+x)dx=I   { ((u=ln(1+x))),((v′=1)) :}⇒  { ((u′=(1/(1+x)))),((v=x)) :}  I = [xln(1+x)]_0 ^1 −∫_0 ^1 (x/(1+x))dx  =ln2−∫_0 ^1 ((x+1−1)/(1+x))dx  =ln2−[x]_0 ^1 +[ln(1+x)]_0 ^1 = 2ln2−1  ⇒ln(L)=U⇒L=e^U   L=e^(ln4−1) =(4/e)..
$$\mathrm{L}=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} ;\:\:\mathrm{U}=\mathrm{ln}\left(\mathrm{L}\right) \\ $$$$\mathrm{U}=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)=\mathrm{f}\left(\frac{\mathrm{k}}{\mathrm{n}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}=\mathrm{I} \\ $$$$\begin{cases}{\mathrm{u}=\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}\\{\mathrm{v}'=\mathrm{1}}\end{cases}\Rightarrow\:\begin{cases}{\mathrm{u}'=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}}\\{\mathrm{v}=\mathrm{x}}\end{cases} \\ $$$$\mathrm{I}\:=\:\left[\mathrm{xln}\left(\mathrm{1}+\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{ln2}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}+\mathrm{1}−\mathrm{1}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{ln2}−\left[\mathrm{x}\right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\:\mathrm{2ln2}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{ln}\left(\mathrm{L}\right)=\mathrm{U}\Rightarrow\mathrm{L}=\mathrm{e}^{\mathrm{U}} \\ $$$$\mathrm{L}=\mathrm{e}^{\mathrm{ln4}−\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{e}}.. \\ $$

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