Friday, January 29, 2016

limits - Evaluate $lim_{n to infty }frac{(n!)^{1/n}}{n}$.











Evaluate
$$\lim_{n \to \infty }\frac{(n!)^{1/n}}{n}.$$



Can anyone help me with this? I have no idea how to start with. Thank you.


Answer



Let's work it out elementarily by wisely applying Cauchy-d'Alembert criterion:




$$\lim_{n\to\infty} \frac{n!^{\frac{1}{n}}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{(n+1)}}\cdot \frac{n^{n}}{n!} = \lim_{n\to\infty} \frac{n^{n}}{(n+1)^{n}} =\lim_{n\to\infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}}=\frac{1}{e}. $$



Also notice that by applying Stolz–Cesàro theorem you get the celebre limit:



$$\lim_{n\to\infty} (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}} = \frac{1}{e}.$$



The sequence $L_{n} = (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}$ is called Lalescu sequence, after the name of a great Romanian mathematician, Traian Lalescu.



Q.E.D.


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