# Ngô Quốc Anh

## September 17, 2007

### Bất đẳng thức cho dãy Lalescu

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 6:00

Prove that for the T. Lalescu‘s sequence $L_{n}=\sqrt [n+1]{(n+1) !}-\sqrt [n]{n!}$ exists the “clenching$\left(\frac{n}{n+1}\right)^{n+1}< L_{n}<\left(\frac{n}{n+1}\right)^{n}$.

$1\blacktriangleright$ $\mathrm{L.H.S.\ : }$ Here is a nice and dificult problem (if is possible, without derivatives) !
$2\blacktriangleright$ $\mathrm{R.H.S.\ :\ G.M.\le A.M.\ for\ }$
$x_{k}: =\left\{\begin{array}{ccc}\frac{\sqrt [n]{n!}}{n}&\mathrm{if}& k\in\overline{1,n}\\ \\ \left(\frac{n}{n+1}\right)^{n}&\mathrm{if}& k = n+1\end{array}\right\|$
$\frac{n\cdot\frac{\sqrt [n]{n!}}{n}+\left(\frac{n}{n+1}\right)^{n}}{n+1}>$ $\sqrt [n+1]{\left(\frac{\sqrt [n]{n!}}{n}\right)^{n}\cdot\left(\frac{n}{n+1}\right)^{n}}=$

$\frac{\sqrt [n+1]{(n+1)!}}{n+1}\implies$ $\sqrt [n]{n!}+\left(\frac{n}{n+1}\right)^{n}>\sqrt [n+1]{(n+1)!}\implies$ $\sqrt [n+1]{(n+1)!}-\sqrt [n]{n!}<\left(\frac{n}{n+1}\right)^{n}$ $\implies$ $L_{n}<\left(\frac{n}{n+1}\right)^{n}$.

Remark.
$\lim_{n\to\infty }\left(\frac{n}{n+1}\right)^{n}=\frac{1}{e}\implies\boxed{\ \lim_{n\to\infty}\left(\sqrt [n+1]{(n+1)!}-\sqrt [n]{n!}\right)=\frac{1}{e}\ }$.