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This question already has an answer here:

Find $\lim_{n\to \infty}((n+1)!)^{\frac{1}{n+1}}-((n)!)^{\frac{1}{n}}.$

We need to deal the limit $\lim_{n\to \infty} \frac{\log(1)+\log(2)+...+\log(n)}{n}$. We know that $\lim_{n\to \infty} \log(n)=\infty \implies \lim_{n\to \infty} \frac{\log(1)+\log(2)+...+\log(n)}{n}=\infty$(since, By Cauchy's first theorem on limit). Hence we get $\infty-\infty$. How do I show that there exists finite limit?

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marked as duplicate by Paramanand Singh limits Dec 6 '18 at 16:10

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  • $\begingroup$ Let me check. But the limit is infinity. $\endgroup$ – Unknown x Dec 6 '18 at 16:01
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    $\begingroup$ The limit should be $1/e$. $\endgroup$ – Paramanand Singh Dec 6 '18 at 16:04
  • $\begingroup$ How do we get?Can you give some hint? $\endgroup$ – Unknown x Dec 6 '18 at 16:05
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HINT:

Using Stirling's Formula we have

$$\begin{align} \left((n+1)!\right)^{1/(n+1)}-\left(n!\right)^{1/n}&=\left(\left(\sqrt{2\pi(n+1)}\left(\frac{n+1}{e}\right)^{n+1}\right)\left(1+O(1/n)\right)\right)^{1/(n+1)}\\\\ &-\left(\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right)\left(1+O(1/n)\right)\right)^{1/n} \end{align}$$

Can you finish now?


Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.

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  • $\begingroup$ This it is not convincing at all. Did you really check that this gets somewhere ? $\endgroup$ – Ewan Delanoy Dec 6 '18 at 16:13
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    $\begingroup$ @EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$. $\endgroup$ – Mark Viola Dec 6 '18 at 16:15
  • $\begingroup$ @EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER. $\endgroup$ – Mark Viola Dec 6 '18 at 16:22
  • $\begingroup$ Is your sirling formula correct? $\sqrt{2 \pi n}$ in the numerator? $\endgroup$ – Unknown x Dec 6 '18 at 17:23
  • $\begingroup$ your alternate approach is very excellent :) $\endgroup$ – Unknown x Dec 6 '18 at 17:25

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