What is the simple demonstration with elementary means for Lalescu Sequence: $$\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) = \frac{1}{e}?$$
(Traian Lalescu-romanian mathematician (1882-1929))
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1Stirling approximation is a good way. – Claude Leibovici Apr 17 '14 at 13:56
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For your curiosity, it is possible to show that, for large values of $n$, $\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\simeq \frac{1}{e}+\frac{1}{2 e n}$ – Claude Leibovici Apr 17 '14 at 14:07
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Apply Stolz–Cesàro theorem to $a_n=(n!)^{1/n}$ and $b_n=n$, then the desired limit equals $$ \ell=\lim\limits_{n\to\infty}\frac{(n!)^{1/n}}{n}=\lim\limits_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n} $$ Now recall one more formula $$ \lim\limits_{n\to\infty}|c_n|^{1/n}=\lim\limits_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| $$ Taking $c_n=n!/n^n$ we get $$ \ell =\lim\limits_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!} =\lim\limits_{n\to\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e} $$
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The way you use a reciprocal Stolz-Cesaro lemma, which should be demonstrated first! – medicu Apr 17 '14 at 14:09