# Limit of the sequence $\lim_{n\to\infty}(n!)^{1/n^2}$ [closed]

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

Thank you!

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## closed as off-topic by Harish Chandra Rajpoot, S.Panja-1729, Claude Leibovici, MATHEMATIKER, ServaesSep 26 '15 at 12:07

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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
There are several votes to close this as off-topic. Related discussion on meta: Why close old questions with accepted answers using the “no context” reason? – Martin Sleziak Sep 26 '15 at 8:52

use $1<n!<n^n$.and $\displaystyle \lim_{n\to\infty}n^{\frac{1}{n}}=1$

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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
I will add that the limit you mention is subject of several posts on this site. For example, math.stackexchange.com/questions/28348/… or math.stackexchange.com/questions/115822/… (and many other posts). – Martin Sleziak Sep 26 '15 at 8:40

By Stirling formula we have $$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$ so $$(n!)^{1/n^2}\sim_\infty \left(\frac{n}{e}\right)^{1/n}(2\pi n)^{1/2n^2}\to_\infty1$$

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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 $L=1$

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