Limit $\lim\limits_{n\to\infty}((n-1)!)^{\frac{1}{n}}$

Question

I try to find the $\lim_{n \to \infty} ((n - 1)!)^{\frac{1}{n}}$ using Stolz-Cesaro convergece theorem and it goes like this: $L = \lim_{n \to \infty} \frac{n!}{(n-1)!} = \infty$. Am I right?

Answer 1

Hint

$$\ln \left( \sqrt[n]{(n - 1)!}\right)=\frac{\ln(1)+\ln(2)+..\ln(n-1)}{n} \,.$$

Now if you use S-C you get exactly what you said.

I guess you used a consequence of SC, which says:

C: If $a_n >0$ and $\lim_n \frac{a_n}{a_{n-1}}$ exists and is equal to $l$ then $$\lim_n \sqrt[n]{a_n}=l$$

Answer 2

use Stolz-Cesaro $$I=\lim_{n\to\infty}((n-1)!)^{\frac{1}{n}}=\lim_{n\to\infty}e^{\dfrac{\ln{(n-1)!}}{n}}=e^{\lim_{n\to\infty}\dfrac{\ln{(n!)}-\ln{(n-1)!}}{(n+1)-n}}=e^{\lim_{n\to\infty}\ln{n}}\to\infty$$

In fact,it is very know $$n!>(\dfrac{n}{3})^n$$ so $$((n-1)!)^{\frac{1}{n}}>\left(\dfrac{n-1}{3}\right)^{\frac{n-1}{n}}\to\infty$$

Answer 3

There are other ways to compute this kind of limits. They involve Stirling approximations for n! [http://en.wikipedia.org/wiki/Stirling%27s_approximation]. Since you are concerned by almost infinite value of n, the simplest could be sufficient : Log[n!] can be approximated using (n Log[n] - n). From there, you would easily show that [(n-1)!]^(1/n) goes to infinity.
You also can use a more complex and more precise formula from Stirling which write n!= Sqrt[2 Pi n] (n / e)^n and reach the same conclusion.