Could anyone help me finding the following limit: $$\lim_{n\to\infty}(n!)^{1/n^2}$$
Thank you!
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1You 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
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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
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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 08:52
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Another post about the same limit: Using the squeeze theorem to determine a limit $\lim_{n\to\infty} (n!)^{\frac{1}{n^2}}$ – Martin Sleziak May 21 '18 at 06:43
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use $1<n!<n^n$.and $\displaystyle \lim_{n\to\infty}n^{\frac{1}{n}}=1$
math110
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1Thank 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
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I will add that the limit you mention is subject of several posts on this site. For example, http://math.stackexchange.com/questions/28348/proof-that-lim-n-rightarrow-infty-sqrtnn-1 or http://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 (and many other posts). – Martin Sleziak Sep 26 '15 at 08:40
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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 $L=1$
Alex
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