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Question Number 116451 by bemath last updated on 04/Oct/20

lim_(n→∞)  ((((n+1)(n+2)(n+3)...(n+n)))^(1/n) /n) =?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{n}}]{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)...\left(\mathrm{n}+\mathrm{n}\right)}}{\mathrm{n}}\:=? \\ $$

Commented by Olaf last updated on 04/Oct/20

Sorry I′m wrong !  I calculated another expression !  !??! I′m tired !

$$\boldsymbol{\mathrm{Sorry}}\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{wrong}}\:! \\ $$$$\mathrm{I}\:\boldsymbol{\mathrm{calculated}}\:\boldsymbol{\mathrm{another}}\:\boldsymbol{\mathrm{expression}}\:! \\ $$$$!??!\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{tired}}\:! \\ $$

Answered by Olaf last updated on 04/Oct/20

u_n  =^n (√((Π_(k=1) ^n (n+k))/n))  v_n = lnu_n   v_n = (1/n)(lnΠ_(k=1) ^n (n+k)−lnn)  v_n = (1/n)(lnΠ_(k=1) ^n n(1+(k/n))−lnn)  v_n = (1/n)(lnn^n Π_(k=1) ^n (1+(k/n))−lnn)  v_n = (1/n)(nlnn+lnΠ_(k=1) ^n (1+(k/n))−lnn)  v_n = (1/n)(nlnn+Σ_(k=1) ^n ln(1+(k/n))−lnn)  ln(1+(k/n)) ∼_∞  (k/n)  v_n ∼_∞  (1/n)(nlnn+Σ_(k=1) ^n (k/n)−lnn)  v_n ∼_∞  (1/n)(nlnn+(1/n).((n(n+1))/2)−lnn)  v_n ∼_∞  lnn+((n+1)/(2n))−((lnn)/n) = +∞+(1/2)−0  lim_(n→∞) v_n  = +∞  lim_(n→∞) u_n  = lim_(n→∞) e^v_n   = +∞

$${u}_{{n}} \:=\:^{{n}} \sqrt{\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({n}+{k}\right)}{{n}}} \\ $$$${v}_{{n}} =\:\mathrm{ln}{u}_{{n}} \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\mathrm{ln}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({n}+{k}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\mathrm{ln}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{n}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\mathrm{ln}{n}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\mathrm{ln}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}}\right)−\mathrm{ln}{n}\right) \\ $$$$\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}}\right)\:\underset{\infty} {\sim}\:\frac{{k}}{{n}} \\ $$$${v}_{{n}} \underset{\infty} {\sim}\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{{n}}−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} \underset{\infty} {\sim}\:\frac{\mathrm{1}}{{n}}\left({n}\mathrm{ln}{n}+\frac{\mathrm{1}}{{n}}.\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{ln}{n}\right) \\ $$$${v}_{{n}} \underset{\infty} {\sim}\:\mathrm{ln}{n}+\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{ln}{n}}{{n}}\:=\:+\infty+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{v}_{{n}} \:=\:+\infty \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{{v}_{{n}} } \:=\:+\infty \\ $$$$ \\ $$$$ \\ $$

Commented by bemath last updated on 04/Oct/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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)

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{n}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{n}}\right)\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}\right)..}=\mathrm{y} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)=\mathrm{logy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}=\mathrm{logy} \\ $$$$\left[\mathrm{xlog}\left(\mathrm{1}+\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}=\mathrm{logy} \\ $$$$\mathrm{log}\left(\mathrm{2}\right)−\mathrm{1}+\left[\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{logy} \\ $$$$\mathrm{2log}\left(\mathrm{2}\right)−\mathrm{1}=\mathrm{logy} \\ $$$$\mathrm{y}=\frac{\mathrm{4}}{\mathrm{e}} \\ $$

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...

$$\:{solution} \\ $$$$\:\:{l}={lim}_{{n}\rightarrow\infty} \frac{\sqrt[{{n}}]{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{n}\right)}}{{n}} \\ $$$$\:\:{l}={lim}_{{n}\rightarrow\infty\:} \sqrt[{{n}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)...\left(\mathrm{1}+\frac{{n}}{{n}}\right)} \\ $$$${log}\left({l}\right)\:={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\:\underset{{k}=} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+\frac{{k}}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {log}\left(\mathrm{1}+{x}\right){dx}\:=\int_{\mathrm{1}\:\:} ^{\:\mathrm{2}} {log}\left({t}\right){dt} \\ $$$$=\:\left[{t}\:{log}\left({t}\right)\:−{t}\:\right]_{\mathrm{1}} ^{\mathrm{2}} =\:\mathrm{2}{log}\left(\mathrm{2}\right)\:−\mathrm{2}+\mathrm{1} \\ $$$$\:\:=\:{log}\left(\mathrm{4}\right)−{log}\left({e}\right)\:=\:{log}\left(\frac{\mathrm{4}}{{e}}\right) \\ $$$$\:\:\:\therefore\:\:\:\:\:\:{l}\::=\:\frac{\mathrm{4}}{{e}}\:\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\mathrm{1970}... \\ $$$$ \\ $$$$ \\ $$

Commented by bemath last updated on 04/Oct/20

yes...thank you

$$\mathrm{yes}...\mathrm{thank}\:\mathrm{you}\: \\ $$

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