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The question is as in the topic.

$$ \lim_{n \rightarrow \infty} \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} $$

I can't think of an integral it may represent. Any help appreciated.

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Alternately, one might use Stirling's approximation. – Lucian May 20 '14 at 5:22
up vote 9 down vote accepted

Taking the log gives $$-\ln n+{1\over n}\big(\ln (n+1)+\cdots +\ln (2n)\big)={1\over n}\left([\ln(n+1)-\ln n]+\cdots+[\ln(2n)-\ln n]\right)$$ $$={1\over n}\left(\ln\left(1+{1\over n}\right)+\cdots+\ln\left(1+{n\over n}\right)\right)\to\int_1^2\ln x\,\mathrm{d}x=2\ln 2-1$$

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I don't get it. Could you explain me how do we go from the factorial into a logarithm? I didn't have anything like that neither in my lectures, nor in my classes. – Mateusz May 19 '14 at 22:22
@Mateusz $\ln (n!)=\ln(1\cdot 2\cdots n)=\ln 1+\ln 2+\cdots+\ln n$. – NotNotLogical May 19 '14 at 22:23
There was no question... – Mateusz May 19 '14 at 22:24
Out of curiosity, how did you come up with this? Was there a particular aspect of the problem that led you to try this? – templatetypedef May 20 '14 at 0:39
@templatetypedef I have used the "trick" that a logarithm of a factorial becomes a sum many times, so that was one thing. Also something about the limit seemed like L'hopital's rule, so I started thinking about using either $e^x$ or $\ln x$. And then, I have to admit, it just fell into place - pretty lucky there :) – NotNotLogical May 20 '14 at 0:46

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