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

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

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.

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