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Question Number 145783 by SOMEDAVONG last updated on 08/Jul/21

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