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Question Number 122241 by TANMAY PANACEA last updated on 15/Nov/20
limit...
$${limit}… \\ $$
Commented by TANMAY PANACEA last updated on 15/Nov/20
Commented by Dwaipayan Shikari last updated on 15/Nov/20
(((n!))^(1/n) /n).(((Σ^∞ (1/n)))/(log(n))) lim_(n→∞) ((((n!)/n^n ))^(1/n) =(((n^n /n^n ).(1/e^n ).(√(2πn))))^(1/n) ).(((Σ^∞ (1/n)−logn+logn)/(logn))) =lim_(n→∞) (1/e)((γ/(logn))+1) (γ=Eulerian constant) =(1/e)
$$\frac{\sqrt[{{n}}]{{n}!}}{{n}}.\frac{\left(\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}\right)}{{log}\left({n}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{{n}}]{\frac{{n}!}{{n}^{{n}} }}\:=\sqrt[{{n}}]{\frac{{n}^{{n}} }{{n}^{{n}} }.\frac{\mathrm{1}}{{e}^{{n}} }.\sqrt{\mathrm{2}\pi{n}}}\right).\left(\frac{\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−{logn}+{logn}}{{logn}}\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{1}}{{e}}\left(\frac{\gamma}{{logn}}+\mathrm{1}\right)\:\:\:\:\:\:\left(\gamma=\mathscr{E}{ulerian}\:\:{constant}\right) \\ $$$$=\frac{\mathrm{1}}{{e}} \\ $$
Commented by mnjuly1970 last updated on 15/Nov/20
mercey mr dwaiypayan your work and effort is very admirable. god keep you′′ young sir′′ and good luck.
$${mercey}\:{mr}\:{dwaiypayan} \\ $$$$\:{your}\:{work}\:{and}\:{effort}\: \\ $$$$\:{is}\:{very}\:{admirable}. \\ $$$${god}\:{keep}\:{you}''\:{young}\:{sir}'' \\ $$$${and}\:{good}\:{luck}. \\ $$
Commented by Dwaipayan Shikari last updated on 15/Nov/20
With always pleasure sir! :)
$$\left.{With}\:{always}\:{pleasure}\:{sir}!\::\right) \\ $$
Commented by TANMAY PANACEA last updated on 15/Nov/20
excellent
$${excellent} \\ $$
Commented by TANMAY PANACEA last updated on 15/Nov/20
Commented by TANMAY PANACEA last updated on 15/Nov/20
lim_(n→∞(((n!)/n^n ))^(1/n) ×lim_(n→∞) ((((1/1)+(1/2)+(1/3)+..+(1/n)−log_e n+log_e n))/(log_e n))) First limit lim_(n→∞) (((n!)/n^n ))^(1/n) T_n =((n!)/n^n ) so T_(n+1) =(((n+1)!)/((n+1)^(n+1) )) lim_(n→∞) (T_(n+1) /T_n )=(((n+1)n!)/((n+1)(n+1)^n ))×(n^n /(n!))=((n/(n+1)))^n =((1/(1+(1/n))))^n letP=(1/n) so lim_(y→0) ((1/(1+y)))^(1/y) =lim_(y→0) (1/((1+y)^(1/y) )) log_e p=−lim_(y→0) ((log_e (1+y))/y)=−1 p=e^(−1) second limit lim_(n→∞) ((((1/1)+(1/2)+(1/3)+..+(1/n)−log_e n+log_e n)/(log_e n))) (1+(γ/(log_e n)))→(1+(γ/∞))=1 so answet=(1/e)×1=(1/e)
$${li}\underset{{n}\rightarrow\infty\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} ×{li}\underset{{n}\rightarrow\infty} {{m}}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+..+\frac{\mathrm{1}}{{n}}−{log}_{{e}} {n}+{log}_{{e}} {n}\right)}{{log}_{{e}} {n}}} {{m}} \\ $$$${First}\:{limit}\:\:{li}\underset{{n}\rightarrow\infty} {{m}}\:\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$${T}_{{n}} =\frac{{n}!}{{n}^{{n}} }\:\:\:{so}\:{T}_{{n}+\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\:\frac{{T}_{{n}+\mathrm{1}} }{{T}_{{n}} }=\frac{\left({n}+\mathrm{1}\right){n}!}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{{n}} }×\frac{{n}^{{n}} }{{n}!}=\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{{n}} =\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\right)^{{n}} \\ $$$${letP}=\frac{\mathrm{1}}{{n}}\:\:\:\:{so}\:\:\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{1}+{y}}\right)^{\frac{\mathrm{1}}{{y}}} ={li}\underset{{y}\rightarrow\mathrm{0}} {{m}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+{y}\right)^{\frac{\mathrm{1}}{{y}}} } \\ $$$${log}_{{e}} {p}=−{li}\underset{{y}\rightarrow\mathrm{0}} {{m}}\:\frac{{log}_{{e}} \left(\mathrm{1}+{y}\right)}{{y}}=−\mathrm{1} \\ $$$${p}={e}^{−\mathrm{1}} \\ $$$${second}\:{limit} \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\left(\frac{\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+..+\frac{\mathrm{1}}{{n}}−{log}_{{e}} {n}+{log}_{{e}} {n}}{{log}_{{e}} {n}}\right) \\ $$$$\left(\mathrm{1}+\frac{\gamma}{{log}_{{e}} {n}}\right)\rightarrow\left(\mathrm{1}+\frac{\gamma}{\infty}\right)=\mathrm{1} \\ $$$${so}\:{answet}=\frac{\mathrm{1}}{{e}}×\mathrm{1}=\frac{\mathrm{1}}{{e}} \\ $$
Commented by mnjuly1970 last updated on 15/Nov/20
excellent.thank you so much ′′sir tanmay′′...
$${excellent}.{thank}\:{you}\:{so}\:{much}\:''{sir}\:{tanmay}''… \\ $$
Commented by mathmax by abdo last updated on 15/Nov/20
let U_n =(((n!)^(1/n) )/(n log(n)))(1+(1/2)+...+(1/n)) ⇒U_n =(((n!)^(1/n) )/(nlog(n)))H_n we have H_n ∼ln(n)+γ and n! ∼n^n e^(−n) (√(2πn)) ⇒ U_n ∼(((n^n e^(−n) (√(2πn)))^(1/n) )/(nlog(n)))(logn +γ) =((n e^(−1) (2πn)^(1/(2n)) )/(nlog(n)))(γ +logn) =e^(−1) (2πn)^(1/(2n)) (1+(γ/(logn))) we have lim_(n→+∞) (2πn)^(1/(2n)) =lim_(n→+∞) e^((log(2πn))/(2n)) =e^o =1 ⇒ lim_(n→+∞) U_n =(1/e)
$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\frac{\left(\mathrm{n}!\right)^{\frac{\mathrm{1}}{\mathrm{n}}} }{\mathrm{n}\:\mathrm{log}\left(\mathrm{n}\right)}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{n}}\right)\:\Rightarrow\mathrm{U}_{\mathrm{n}} =\frac{\left(\mathrm{n}!\right)^{\frac{\mathrm{1}}{\mathrm{n}}} }{\mathrm{nlog}\left(\mathrm{n}\right)}\mathrm{H}_{\mathrm{n}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{H}_{\mathrm{n}} \sim\mathrm{ln}\left(\mathrm{n}\right)+\gamma\:\:\mathrm{and}\:\mathrm{n}!\:\sim\mathrm{n}^{\mathrm{n}} \mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} \sim\frac{\left(\mathrm{n}^{\mathrm{n}} \mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} }{\mathrm{nlog}\left(\mathrm{n}\right)}\left(\mathrm{logn}\:+\gamma\right) \\ $$$$=\frac{\mathrm{n}\:\mathrm{e}^{−\mathrm{1}} \left(\mathrm{2}\pi\mathrm{n}\right)^{\frac{\mathrm{1}}{\mathrm{2n}}} }{\mathrm{nlog}\left(\mathrm{n}\right)}\left(\gamma\:+\mathrm{logn}\right)\:=\mathrm{e}^{−\mathrm{1}} \left(\mathrm{2}\pi\mathrm{n}\right)^{\frac{\mathrm{1}}{\mathrm{2n}}} \left(\mathrm{1}+\frac{\gamma}{\mathrm{logn}}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \left(\mathrm{2}\pi\mathrm{n}\right)^{\frac{\mathrm{1}}{\mathrm{2n}}} \:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{e}^{\frac{\mathrm{log}\left(\mathrm{2}\pi\mathrm{n}\right)}{\mathrm{2n}}} \:=\mathrm{e}^{\mathrm{o}} \:=\mathrm{1}\:\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$$$ \\ $$

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