194. The Lalescu's sequence. Properties.

by Virgil Nicula, Dec 22, 2010, 7:05 AM

The sequence $L_n=\sqrt [n+1]{(n+1)!}-\sqrt [n]{n!}$ , where $n\in\mathbb N$ is named Lalescu's sequence. Prove that $L_n\rightarrow\frac 1e$ .

Proof. Prove easily that $L_n=a_n\cdot b_n\cdot c_n$ , where $a_n=\frac {\sqrt [n]{n!}}{n}\rightarrow$ $\frac 1e\ ,\ b_n=\frac {e^{\frac {\ln (n+1)!}{n+1}-\frac {\ln n!}{n}}-1}{\frac {\ln (n+1)!}{n+1}-\frac {\ln n!}{n}}$ $\rightarrow 1\ ,\ c_n=\frac {n\cdot\ln (n+1)!-(n+1)\cdot\ln n!}{n+1}\rightarrow 1$ .

Indeed, for the sequence $c_n=\frac {n\cdot\ln (n+1)!-(n+1)\cdot\ln n!}{n+1}= \frac {n\cdot\ln (n+1)-\ln n!}{n+1}\equiv\frac {u_n}{v_n}$ using the Cesaro-Stolz's lemma obtain that $\frac {u_{n+1}-u_n}{v_{n+1}-v_n}=$

$\frac {(n+1)\cdot\ln (n+2)-n\cdot\ln (n+1)-\ln (n+1)!+\ln n!}{(n+2)-(n+1)}=$ $(n+1)\cdot\ln (n+2)-n\cdot\ln (n+1)-\ln (n+1)=$ $(n+1)\cdot\ln\frac {n+2}{n+1}=$

$\ln\left(1+\frac {1}{n+1}\right)^{n+1}\rightarrow 1$ . Therefore and $\frac {u_n}{v_n}\rightarrow 1$ , i.e. $c_n\rightarrow 1$ . In conclusion, $L_n\rightarrow\frac 1e$ .

Remark. $\boxed{\ \left(\frac {n}{n+1}\right)^{n+1}\ <\ \sqrt [n+1]{(n+1)!}-\sqrt [n]{n!}\ <\ \left(\frac {n}{n+1}\right)^n\ }$ . In conclusion $L_n\rightarrow\frac 1e$ .

For the right hand side (RHS) exists an elementary nice proof (own !). Apply the remarkable inequality $\mathrm{A.M}\ge\mathrm{ G.A}$ to the numbers

$a_k=\frac {\sqrt [n]{n!}}{n}$ , where $k\in\overline{1,n}$ and $a_{n+1}=\left(\frac {n}{n+1}\right)^n\ :\ \frac {n\cdot \frac {\sqrt [n]{n!}}{n}+\left(\frac {n}{n+1}\right)^n}{n+1}\ge\sqrt [n+1]{\left(\frac {\sqrt [n]{n!}}{n}\right)^n\cdot \left(\frac {n}{n+1}\right)^n}$ $\implies$

$\sqrt [n]{n!}+\left(\frac {n}{n+1}\right)^n\ge \sqrt [n+1]{(n+1)^{n+1}\left(\frac {\sqrt [n]{n!}}{n}\right)^n\cdot \left(\frac {n}{n+1}\right)^n}=$ $\sqrt [n+1]{(n+1)!}\ \implies\ L_n<\left(\frac {n}{n+1}\right)^n$ .
This post has been edited 13 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:33 PM

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Hello Professor Nicula!
At the double inequality, the inequality on the right was demonstrated (even with AM-GM - I don't know if even as above) by Al.Lupas and A. Vernescu in G.M.
But inequality on the left. Who discovered it? Is there an elementary proof - possibly with the AM-GM inequality of this inequality?
(Intreaba : Dorin Marghidanu)

by Dorin, Sep 26, 2020, 9:21 AM

Own problems or extensions/generalizations of some problems which was posted here.

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