1

I need to prove the limit of: $$\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0 $$

I have tried bounding it by a pairwise bigger sequence but I cannot find one that is suitable. I also cannot figure out an $\epsilon$-proof, since I don't know how to adequately handle the factorial to get a substitution for n such that the convergence criterion is fulfilled.

lthz
  • 161

1 Answers1

3

Consider the following: $$ n! = n(n-1)(n-2)\cdots 3 \cdot 2 \cdot 1 \geq \underbrace{{n\over 2}\cdot {n\over 2}\cdots {n \over 2}}_{{n \over 2}\text{times}} = \left({n\over 2}\right)^{n\over 2} \iff \\ \iff \sqrt[n]{n!} \ge \left({\left({n\over 2}\right)^{n\over 2}}\right)^{1\over n} = \sqrt{n\over 2} \iff {1\over \sqrt[n]{n!}} \le \sqrt{2\over n} $$

Also $x_n \ge 0$, so by squeeze theorem: $$ 0 \le \lim_{n\to\infty} {1\over \sqrt[n]{n!}} \le \lim_{n\to \infty} \sqrt{2\over n} = 0 $$

Another way would be Stirling's approximation. Recall: $$ n! \sim \sqrt{2\pi n} \left(n\over e\right)^{n} $$ So: $$ \lim_{n\to\infty} {1\over \sqrt[n]{n!}} = \lim_{n\to\infty} {1\over \sqrt[n]{2\pi n} }\left(e\over n\right) = 1\cdot 0 = 0 $$

roman
  • 5,391
  • Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative. – lthz Dec 18 '18 at 15:53