lim_(n→∞) [(((n+1)(n+2)......3n)/n^(2n) )]^(1/n)


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Question Number 100904 by Dwaipayan Shikari last updated on 29/Jun/20

$li \underset{n \to \infty}{m} {[\frac{(n + 1) (n + 2) . . . . . . 3 n}{n^{2 n}}]}^{\frac{1}{n}}$
Commented by Ar Brandon last updated on 29/Jun/20
I like how you transformed the sum directly into integral without decomposing it into simpler sums.��
Commented by Dwaipayan Shikari last updated on 29/Jun/20

$s u p p o s e y = l i \underset{n \to \infty}{m} {[\frac{(n + 1) (n + 2) . . . . 3 n}{n . n . . . . . . 2 n t i m e s}]}^{\frac{1}{n}}$ $l o g y = \frac{1}{n} l i \underset{n \to \infty}{m} [l o g (1 + \frac{1}{n}) + l o g (1 + \frac{2}{n}) + . . . . + l o g (1 + \frac{2 n}{n})]$ $l o g y = \frac{1}{n} l i \underset{n \to \infty}{m} \underset{r = 1}{\sum^{2 n}} l o g (1 + \frac{r}{n})$ $l o g y = \int_{0}^{2} l o g (1 + x) d x$ $l o g y = {[x l o g (1 + x)]}_{0}^{2} - \int_{0}^{2} \frac{x}{x + 1}$ $l o g y = 2 l o g 3 - {[x - l o g (x + 1)]}_{0}^{2}$ $l o g y = 2 l o g 3 - 2 + l o g 3$ $l o g y = 3 l o g 3 - 2$ $y = e^{3 l o g 3 - 2} = \frac{27}{e^{2}}$ $c h e c k t h i s$
Commented by mscbed last updated on 29/Jun/20


	Answered by Ar Brandon last updated on 29/Jun/20
	$l=lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =lim_(n→∞) [(1+(1/n))(1+(2/n))∙∙∙(1+((2n)/n))]^(1/n) ⇒l=lim_(n→∞) Π_(k=1) ^(2n) (1+(k/n))^(1/n) ⇒lnl=lim_(n→∞) ln{Π_(k=1) ^(2n) (1+(k/n))^(1/n) } ⇒lnl=lim_(n→∞) Σ_(k=1) ^(2n) {ln(1+(k/n))^(1/n) }=lim_(n→∞) (1/n)Σ_(k=1) ^(2n) ln(1+(k/n)) =lim_(n→∞) (1/n){Σ_(k=1) ^n ln(1+(k/n))+Σ_(k=n+1) ^(2n) ln(1+(k/n))} =lim_(n→∞) (1/n){Σ_(k=1) ^n ln(1+(k/n))+Σ_(p=1) ^n ln(1+((n+p)/n))} =lim_(n→∞) (1/n)Σ_(k=1) ^n ln(1+(k/n))+lim_(n→∞) (1/n)Σ_(p=1) ^n ln(2+(p/n)) =∫_1 ^2 lnxdx+∫_2 ^3 lnxdx=[x(lnx−1)]_1 ^2 +[x(lnx−1)]_2 ^3 =2(ln2−1)+1+3(ln3−1)−2(ln2−1)=3ln3−2 ⇒l=e^(3ln3−2) =(3^3 /e^2 )⇒lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =((27)/e^2 )$
	$l = \lim_{n \to \infty} {[\frac{(n + 1) (n + 2) . . . (n + 2 n)}{n^{2 n}}]}^{\frac{1}{n}} = \lim_{n \to \infty} {[(1 + \frac{1}{n}) (1 + \frac{2}{n}) \cdot \cdot \cdot (1 + \frac{2 n}{n})]}^{\frac{1}{n}}$ $\Rightarrow l = \lim_{n \to \infty} \underset{k = 1}{\prod^{2 n}} {(1 + \frac{k}{n})}^{\frac{1}{n}} \Rightarrow \ln l = \underset{n \to \infty}{\lim l n} {\underset{k = 1}{\prod^{2 n}} {(1 + \frac{k}{n})}^{\frac{1}{n}}}$ $\Rightarrow \ln l = \lim_{n \to \infty} \underset{k = 1}{\sum^{2 n}} {\ln {(1 + \frac{k}{n})}^{\frac{1}{n}}} = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{2 n}} \ln (1 + \frac{k}{n})$ $= \lim_{n \to \infty} \frac{1}{n} {\underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n}) + \underset{k = n + 1}{\sum^{2 n}} \ln (1 + \frac{k}{n})}$ $= \lim_{n \to \infty} \frac{1}{n} {\underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n}) + \underset{p = 1}{\sum^{n}} \ln (1 + \frac{n + p}{n})}$ $= \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n}) + \lim_{n \to \infty} \frac{1}{n} \underset{p = 1}{\sum^{n}} \ln (2 + \frac{p}{n})$ $= \int_{1}^{2} lnxdx + \int_{2}^{3} lnxdx = {[x (lnx - 1)]}_{1}^{2} + {[x (lnx - 1)]}_{2}^{3}$ $= 2 (\ln 2 - 1) + 1 + 3 (\ln 3 - 1) - 2 (\ln 2 - 1) = 3 \ln 3 - 2$ $\Rightarrow l = e^{3 \ln 3 - 2} = \frac{3^{3}}{e^{2}} \Rightarrow \lim_{n \to \infty} {[\frac{(n + 1) (n + 2) . . . (n + 2 n)}{n^{2 n}}]}^{\frac{1}{n}} = \frac{27}{e^{2}}$
	Answered by mathmax by abdo last updated on 29/Jun/20
	$let A_n ={(((n+1)(n+2)....(n+2n))/n^(2n) )}^(1/n) ⇒lnA_n =(1/n)ln(((Π_(k=1) ^(2n) (n+k))/n^(2n) )) =(1/n){Σ_(k=1) ^(2n) ln(n+k)−2nln(n)}=(1/n)Σ_(k=1) ^(2n) ln (n+k)−2ln(n) =(1/n)Σ_(k=1) ^(2n) ( lnn+ln(1+(k/n)))−2ln(n) =(1/n)Σ_(k=1) ^(2n) ln(1+(k/n)) =(1/n)Σ_(k=1) ^(n−1) ln(1+(k/n)) +(1/n)Σ_(k=n) ^(2n) ln(1+(k/n))(→p=k−n) =(1/n)Σ_(k=1) ^(n−1) ln(1+(k/n)) +(1/n)Σ_(p=0) ^n ln(1+((p+n)/n)) =(1/n)Σ_(k=1) ^(n−1) ln(1+(k/n))+(1/n)Σ_(p=0) ^n ln(2+(p/n)) we have (1/n)Σ_(k=1) ^(n−1) ln(1+(k/n))→∫_0 ^1 ln(1+x)dx =_(1+x=t) ∫_1 ^2 ln(t)dt =[tlnt−t]_1 ^2 =2ln(2)−2+1 =2ln(2)−1 (1/n)Σ_(p=0) ^n ln(2+(p/n))→∫_0 ^1 ln(2+x)dx =_(2+x=t) ∫_2 ^3 ln(t)dt=[tlnt−t]_2 ^3 =3ln(3)−3−2ln(2)+2 =3ln(3)−2ln(2)−1 ⇒ lim_(n→+∞) lnA_n =2ln(2)−1+3ln(3)−2ln(2)−1 =3ln(3)−2 ⇒ lim_(n→+∞) A_n =e^(3ln(3)−2) =9e^(−2) =(9/e^2 )$
	$let A_{n} = {\frac{(n + 1) (n + 2) . . . . (n + 2 n)}{n^{2 n}}}^{\frac{1}{n}} \Rightarrow {lnA}_{n} = \frac{1}{n} \ln (\frac{\prod_{k = 1}^{2 n} (n + k)}{n^{2 n}})$ $= \frac{1}{n} {\sum_{k = 1}^{2 n} \ln (n + k) - 2 nln (n)} = \frac{1}{n} \sum_{k = 1}^{2 n} \ln (n + k) - 2 \ln (n)$ $= \frac{1}{n} \sum_{k = 1}^{2 n} (lnn + \ln (1 + \frac{k}{n})) - 2 \ln (n) = \frac{1}{n} \sum_{k = 1}^{2 n} \ln (1 + \frac{k}{n})$ $= \frac{1}{n} \sum_{k = 1}^{n - 1} \ln (1 + \frac{k}{n}) + \frac{1}{n} \sum_{k = n}^{2 n} \ln (1 + \frac{k}{n}) (\to p = k - n)$ $= \frac{1}{n} \sum_{k = 1}^{n - 1} \ln (1 + \frac{k}{n}) + \frac{1}{n} \sum_{p = 0}^{n} \ln (1 + \frac{p + n}{n})$ $= \frac{1}{n} \sum_{k = 1}^{n - 1} \ln (1 + \frac{k}{n}) + \frac{1}{n} \sum_{p = 0}^{n} \ln (2 + \frac{p}{n}) we have$ $\frac{1}{n} \sum_{k = 1}^{n - 1} \ln (1 + \frac{k}{n}) \to \int_{0}^{1} \ln (1 + x) dx =_{1 + x = t} \int_{1}^{2} \ln (t) dt = {[tlnt - t]}_{1}^{2}$ $= 2 \ln (2) - 2 + 1 = 2 \ln (2) - 1$ $\frac{1}{n} \sum_{p = 0}^{n} \ln (2 + \frac{p}{n}) \to \int_{0}^{1} \ln (2 + x) dx =_{2 + x = t} \int_{2}^{3} \ln (t) dt = {[tlnt - t]}_{2}^{3}$ $= 3 \ln (3) - 3 - 2 \ln (2) + 2 = 3 \ln (3) - 2 \ln (2) - 1 \Rightarrow$ $\lim_{n \to + \infty} {lnA}_{n} = 2 \ln (2) - 1 + 3 \ln (3) - 2 \ln (2) - 1 = 3 \ln (3) - 2 \Rightarrow$ $\lim_{n \to + \infty} A_{n} = e^{3 \ln (3) - 2} = {9 e}^{- 2} = \frac{9}{e^{2}}$
	Commented by mathmax by abdo last updated on 29/Jun/20

	$sorry \lim_{n \to + \infty} A_{n} = 27 e^{- 2} = \frac{27}{e^{2}}$
	Answered by Ar Brandon last updated on 29/Jun/20
	$A_n =lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =lim_(n→∞) [(1+(1/n))(1+(2/n))∙∙∙(1+((2n)/n))]^(1/n) ⇒A_n =lim_(n→∞) Π_(k=1) ^(2n) (1+(k/n))^(1/n) ⇒lnA_n =lim_(n→∞) ln{Π_(k=1) ^(2n) (1+(k/n))^(1/n) } ⇒lnA_n =lim_(n→∞) Σ_(k=1) ^(2n) {ln(1+(k/n))^(1/n) }=lim_(n→∞) (1/n)Σ_(k=1) ^(2n) ln(1+(k/n)) =∫_1 ^3 lnxdx=[x(lnx−1)]_1 ^3 =3(ln3−1)−(ln1−1) =3ln3−3+1=3ln3−2⇒l=e^(3ln3−2) =(3^3 /e^2 ) ⇒lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =((27)/e^2 )$
	$A_{n} = \lim_{n \to \infty} {[\frac{(n + 1) (n + 2) . . . (n + 2 n)}{n^{2 n}}]}^{\frac{1}{n}} = \lim_{n \to \infty} {[(1 + \frac{1}{n}) (1 + \frac{2}{n}) \cdot \cdot \cdot (1 + \frac{2 n}{n})]}^{\frac{1}{n}}$ $\Rightarrow A_{n} = \lim_{n \to \infty} \underset{k = 1}{\prod^{2 n}} {(1 + \frac{k}{n})}^{\frac{1}{n}} \Rightarrow {lnA}_{n} = \underset{n \to \infty}{\lim l n} {\underset{k = 1}{\prod^{2 n}} {(1 + \frac{k}{n})}^{\frac{1}{n}}}$ $\Rightarrow {lnA}_{n} = \lim_{n \to \infty} \underset{k = 1}{\sum^{2 n}} {\ln {(1 + \frac{k}{n})}^{\frac{1}{n}}} = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{2 n}} \ln (1 + \frac{k}{n})$ $= \int_{1}^{3} lnxdx = {[x (lnx - 1)]}_{1}^{3} = 3 (\ln 3 - 1) - (\ln 1 - 1)$ $= 3 \ln 3 - 3 + 1 = 3 \ln 3 - 2 \Rightarrow l = e^{3 \ln 3 - 2} = \frac{3^{3}}{e^{2}}$ $\Rightarrow \lim_{n \to \infty} {[\frac{(n + 1) (n + 2) . . . (n + 2 n)}{n^{2 n}}]}^{\frac{1}{n}} = \frac{27}{e^{2}}$
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