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$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

I want to prove the following limit: $$\lim_{n \to \infty} \left(\frac{1}{n!}\right)^{1/n} = 0$$ Rewriting into exponentials, this becomes $\displaystyle \exp\left(\frac{1}{n}\log\frac{1}{n!}\right) = \exp\left(-\frac{\log{n!}}{n}\right)$, so it suffices to prove that $$\lim_{n \to \infty} \frac{\log{n!}}{n} = \infty$$ How can I proceed to show this is true?


marked as duplicate by Martin Sleziak, Micah, Erick Wong, user17762, Austin Mohr Dec 1 '12 at 9:27

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HINT: Use the fact that

$$\ln n!=\sum_{k=1}^n\ln k\ge\int_1^n\ln x~dx\;.$$

The summation in the middle is the upper Riemann sum for the integral.

  • $\begingroup$ Thanks! Now I can show that $\lim_{n \to \infty} \frac{\int_1^n \ln {x} dx}{n} = \infty$, either by l'Hopital's Rule or by writing out the integral, and the conclusion follows from conservation of inequalities under the limit. $\endgroup$ – Yiteng Nov 30 '12 at 20:58
  • $\begingroup$ @Peter: You’re welcome! I’d probably just evaluate the integral to get $n\ln n-n$, since dividing that by $n$ is so easy. $\endgroup$ – Brian M. Scott Nov 30 '12 at 21:02

$n!$ has at least $\frac {n-1}2$ factors $\ge\frac n 2$. For $n\ge3$, we have $\frac {n-1}2\ge \frac n 3$, hence $$\sqrt[n]{\frac1{n!}}\le \sqrt[n]{\frac1{\left(\frac n2\right)^{\frac{n}3}}}=\frac1{\sqrt[3]{\frac n2}}\to 0$$


$$\lim_{n \to \infty} \left(\frac{1}{n!}\right)^{1/n} =\lim_{n \to \infty}{\left ( \frac{1}{1} \frac{1}{2} \dots \frac{1}{n-1} \frac{1}{n} \right )} ^\frac{1}{n}\leq \lim _ {n\rightarrow \infty}\frac{\sum_ {i=1}^{n} \frac{i}{n}}{n}$$

from AM-GM Now bound the first $ [\sqrt n ]$ from 1 and the next terms from $ \frac{1}{\sqrt n - 1}$

$$\frac{\sum_ {i=1}^{n} \frac{i}{n}}{n}\leq\frac{\sum_ {i=1}^{[\sqrt{n}]} \frac{i}{n}+\sum_ {i=[\sqrt n]+1}^{n}\frac{i}{n} }{n} \leq \frac{ [\sqrt {n }] \cdot 1+ (n-[\sqrt n])\frac{1}{ [\sqrt n]} } {n}\leq $$ $$\frac{ \sqrt {n }+(n+1-\sqrt n)\frac{1}{ \sqrt n-1} } {n}= O(\sqrt n )\frac{1}{n}\rightarrow0$$ which completes the proof.


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