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Question Number 154495 by mathdanisur last updated on 18/Sep/21

if S_n (t) = n^(1-t) ((((n+1)^(2t) )/(((((n+1)!))^(1/(n+1)) )^t )) - (n^(2t) /((((n!))^(1/n) )^t ))) with t>0 then lim_(n→∞) S_n (t) = te^t

$$\mathrm{if}\:\:\mathrm{S}_{\boldsymbol{\mathrm{n}}} \left(\mathrm{t}\right)\:=\:\mathrm{n}^{\mathrm{1}-\boldsymbol{\mathrm{t}}} \:\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}\boldsymbol{\mathrm{t}}} }{\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\right)^{\boldsymbol{\mathrm{t}}} }\:-\:\frac{\mathrm{n}^{\mathrm{2}\boldsymbol{\mathrm{t}}} }{\left(\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\right)^{\boldsymbol{\mathrm{t}}} }\right) \\ $$ $$\mathrm{with}\:\:\mathrm{t}>\mathrm{0} \\ $$ $$\mathrm{then}\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}S}_{\boldsymbol{\mathrm{n}}} \left(\mathrm{t}\right)\:=\:\mathrm{te}^{\boldsymbol{\mathrm{t}}} \\ $$

Answered by Ar Brandon last updated on 18/Sep/21

S_n (t) = n^(1-t) ((((n+1)^(2t) )/(((((n+1)!))^(1/(n+1)) )^t )) - (n^(2t) /((((n!))^(1/n) )^t ))) S_n (t)=n^(1−t) ((((n+1)^(2t) )/((((√(2π(n+1)(((n+1)/e))^(n+1) )))^(1/(n+1)) )^t ))−(n^(2t) /((((√(2πn((n/e))^n )))^(1/n) )^t ))) =(1/n^(t−1) )((((n+1)^(2t) )/((2πe^(−1) (n+1)^(n+2) )^(t/(2(n+1))) ))−(n^(2t) /((2πe^(−1) n^(n+1) )^(t/(2n)) ))) =(1/n^(t−1) )((((n+1)^(2t−((t(n+2))/(2(n+1)))) )/((2πe^(−1) )^(t/(2(n+1))) ))−(n^(2t−((t(n+1))/(2n))) /((2πe^(−1) )^(t/(2n)) ))) =(1/n^(t−1) )((((n+1)^((3tn+2t)/(2(n+1))) )/((2πe^(−1) )^(t/(2(n+1))) ))−(n^((3tn−t)/(2n)) /((2πe^(−1) )^(t/(2n)) )))

$$\mathrm{S}_{\boldsymbol{\mathrm{n}}} \left(\mathrm{t}\right)\:=\:\mathrm{n}^{\mathrm{1}-\boldsymbol{\mathrm{t}}} \:\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}\boldsymbol{\mathrm{t}}} }{\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\right)^{\boldsymbol{\mathrm{t}}} }\:-\:\frac{\mathrm{n}^{\mathrm{2}\boldsymbol{\mathrm{t}}} }{\left(\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\right)^{\boldsymbol{\mathrm{t}}} }\right) \\ $$ $${S}_{{n}} \left({t}\right)={n}^{\mathrm{1}−{t}} \left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}{t}} }{\left(\sqrt[{{n}+\mathrm{1}}]{\sqrt{\mathrm{2}\pi\left({n}+\mathrm{1}\right)\left(\frac{{n}+\mathrm{1}}{{e}}\right)^{{n}+\mathrm{1}} }}\right)^{{t}} }−\frac{{n}^{\mathrm{2}{t}} }{\left(\sqrt[{{n}}]{\sqrt{\mathrm{2}\pi{n}\left(\frac{{n}}{{e}}\right)^{{n}} }}\right)^{{t}} }\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{n}^{{t}−\mathrm{1}} }\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}{t}} }{\left(\mathrm{2}\pi{e}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)^{{n}+\mathrm{2}} \right)^{\frac{{t}}{\mathrm{2}\left({n}+\mathrm{1}\right)}} }−\frac{{n}^{\mathrm{2}{t}} }{\left(\mathrm{2}\pi{e}^{−\mathrm{1}} {n}^{{n}+\mathrm{1}} \right)^{\frac{{t}}{\mathrm{2}{n}}} }\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{n}^{{t}−\mathrm{1}} }\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}{t}−\frac{{t}\left({n}+\mathrm{2}\right)}{\mathrm{2}\left({n}+\mathrm{1}\right)}} }{\left(\mathrm{2}\pi{e}^{−\mathrm{1}} \right)^{\frac{{t}}{\mathrm{2}\left({n}+\mathrm{1}\right)}} }−\frac{{n}^{\mathrm{2}{t}−\frac{{t}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}}} }{\left(\mathrm{2}\pi{e}^{−\mathrm{1}} \right)^{\frac{{t}}{\mathrm{2}{n}}} }\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{n}^{{t}−\mathrm{1}} }\left(\frac{\left({n}+\mathrm{1}\right)^{\frac{\mathrm{3}{tn}+\mathrm{2}{t}}{\mathrm{2}\left({n}+\mathrm{1}\right)}} }{\left(\mathrm{2}\pi{e}^{−\mathrm{1}} \right)^{\frac{{t}}{\mathrm{2}\left({n}+\mathrm{1}\right)}} }−\frac{{n}^{\frac{\mathrm{3}{tn}−{t}}{\mathrm{2}{n}}} }{\left(\mathrm{2}\pi{e}^{−\mathrm{1}} \right)^{\frac{{t}}{\mathrm{2}{n}}} }\right) \\ $$

Commented bymathdanisur last updated on 19/Sep/21

Thank You Ser, answer = 1

$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{answer}\:=\:\mathrm{1} \\ $$

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