Question Number 116451 by bemath last updated on 04/Oct/20
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{n}}]{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)…\left(\mathrm{n}+\mathrm{n}\right)}}{\mathrm{n}}\:=? \\ $$
Commented by Olaf last updated on 04/Oct/20
$$\boldsymbol{\mathrm{Sorry}}\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{wrong}}\:! \\ $$$$\mathrm{I}\:\boldsymbol{\mathrm{calculated}}\:\boldsymbol{\mathrm{another}}\:\boldsymbol{\mathrm{expression}}\:! \\ $$$$!??!\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{tired}}\:! \\ $$
Answered by Olaf last updated on 04/Oct/20
$${u}_{{n}} \:=\:^{{n}} \sqrt{\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({n}+{k}\right)}{{n}}} \\ $$$${v}_{{n}} =\:\mathrm{ln}{u}_{{n}} \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\mathrm{ln}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({n}+{k}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\mathrm{ln}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{n}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\mathrm{ln}{n}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\mathrm{ln}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$$\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}}\right)\:\underset{\infty} {\sim}\:\frac{{k}}{{n}} \\ $$$${v}_{{n}} \underset{\infty} {\sim}\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{{n}}−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} \underset{\infty} {\sim}\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\frac{\mathrm{1}}{{n}}.\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} \underset{\infty} {\sim}\:\mathrm{ln}{n}+\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{ln}{n}}{{n}}\:=\:+\infty+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{v}_{{n}} \:=\:+\infty \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{{v}_{{n}} } \:=\:+\infty \\ $$$$ \\ $$$$ \\ $$
Commented by bemath last updated on 04/Oct/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Dwaipayan Shikari last updated on 04/Oct/20
![lim_(n→∞) (((1+(1/n))(1+(2/n))(1+(3/n))..))^(1/n) =y lim_(n→∞) (1/n)Σ_(k=1) ^n log(1+(k/n))=logy ∫_0 ^1 log(1+x)dx=logy [xlog(1+x)]_0 ^1 −∫_0 ^1 (x/(1+x))=logy log(2)−1+[log(1+x)]_0 ^1 =logy 2log(2)−1=logy y=(4/e)](https://www.tinkutara.com/question/Q116472.png)
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{n}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{n}}\right)\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}\right)..}=\mathrm{y} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)=\mathrm{logy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}=\mathrm{logy} \\ $$$$\left[\mathrm{xlog}\left(\mathrm{1}+\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}=\mathrm{logy} \\ $$$$\mathrm{log}\left(\mathrm{2}\right)−\mathrm{1}+\left[\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{logy} \\ $$$$\mathrm{2log}\left(\mathrm{2}\right)−\mathrm{1}=\mathrm{logy} \\ $$$$\mathrm{y}=\frac{\mathrm{4}}{\mathrm{e}} \\ $$
Answered by mnjuly1970 last updated on 04/Oct/20
![solution l=lim_(n→∞) ((((n+1)(n+2)...(n+n)))^(1/n) /n) l=lim_(n→∞ ) (((1+(1/n))(1+(2/n))...(1+(n/n))))^(1/n) log(l) =lim_(n→∞) (1/n) Σ_(k=) ^n log(1+(k/n)) = ∫_0 ^( 1) log(1+x)dx =∫_(1 ) ^( 2) log(t)dt = [t log(t) −t ]_1 ^2 = 2log(2) −2+1 = log(4)−log(e) = log((4/e)) ∴ l := (4/e) ✓✓ ...m.n.1970...](https://www.tinkutara.com/question/Q116487.png)
$$\:{solution} \\ $$$$\:\:{l}={lim}_{{n}\rightarrow\infty} \frac{\sqrt[{{n}}]{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left({n}+{n}\right)}}{{n}} \\ $$$$\:\:{l}={lim}_{{n}\rightarrow\infty\:} \sqrt[{{n}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)…\left(\mathrm{1}+\frac{{n}}{{n}}\right)} \\ $$$${log}\left({l}\right)\:={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\:\underset{{k}=} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+\frac{{k}}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {log}\left(\mathrm{1}+{x}\right){dx}\:=\int_{\mathrm{1}\:\:} ^{\:\mathrm{2}} {log}\left({t}\right){dt} \\ $$$$=\:\left[{t}\:{log}\left({t}\right)\:−{t}\:\right]_{\mathrm{1}} ^{\mathrm{2}} =\:\mathrm{2}{log}\left(\mathrm{2}\right)\:−\mathrm{2}+\mathrm{1} \\ $$$$\:\:=\:{log}\left(\mathrm{4}\right)−{log}\left({e}\right)\:=\:{log}\left(\frac{\mathrm{4}}{{e}}\right) \\ $$$$\:\:\:\therefore\:\:\:\:\:\:{l}\::=\:\frac{\mathrm{4}}{{e}}\:\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}.\mathrm{1970}… \\ $$$$ \\ $$$$ \\ $$
Commented by bemath last updated on 04/Oct/20
$$\mathrm{yes}…\mathrm{thank}\:\mathrm{you}\: \\ $$