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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, 91500, Servaes Sep 26 '15 at 12:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Harish Chandra Rajpoot, S.Panja-1729, Claude Leibovici, 91500, Servaes
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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,… or… (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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