# Limit of a sequence

Could anyone help me finding the following limit: $$\lim_{n\to\infty}(n!)^{1/n^2}$$

Thank you!

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You will get a better response if you show what effort you have put into trying to solve the problem yourself. –  Lewy Apr 12 '13 at 15:49
do you know the limit of $M^\frac {1} {n}$ where $M$ is a finite number and $n^\frac {1} {n}$? –  Mathematician Apr 12 '13 at 15:56

use $1<n!<n^n$.and $\displaystyle \lim_{n\to\infty}n^{\frac{1}{n}}=1$
Thank you. I was looking to use something like this, but was wondering what would be the upper bound for $n!$ that I should use. –  John Apr 12 '13 at 16:04
By Stirling formula we have $$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$ so $$(n!)^{1/n^2}\sim_\infty \left(\frac{n}{e}\right)^{1/n}(2\pi n)^{1/2n^2}\to_\infty1$$
Try this ($L$ is the original limit) by first logging the function: $$\lim_{n \to \infty} \frac{\log n!}{n^2}= \lim_{n \to \infty}\frac{\sum_{k=1}^{n} \log k}{n^2} \leq \lim_{n \to \infty} \frac{n \log n}{n^2}=0$$ Hence the limit of the upper bound is $e^0=1$. Now take the lower bound: $$\lim_{n \to \infty}\frac{\sum_{k=1}^{n} \log k}{n^2} \geq \lim_{n \to \infty}\frac{n \log 1}{n^2}=0$$ Hence the limit of the lower bound is $e^0=1$. By the squeeze lemma $L=1$