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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?

1 Answers1

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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.

Mark Viola
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