12

I want to prove the following limit:

$$\lim_{n \to \infty} \sqrt[n+1\;] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}.$$

I searched the forum & found the link here: If $\frac{p_{n+1}}{np_n} \to p > 0 $, then $\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_{n}} \to \frac{p}{e}$ .

But still, there is no way out of the problem. So, please solve it.

Community
  • 1
user92360
  • 975

4 Answers4

8

Hint: Set $p_n = n!$.

Then what is $$\lim_{n \rightarrow \infty}\frac{p_{n+1}}{np_n}?$$

Can you take it from here?

Jose Arnaldo Bebita Dris
  • 1
  • Jose Arnaldo Bebita Dris : The limit you have mentioned is clearly is $1$ . But how it would help me??? ... I didn't know about that "Sterling Approximation"... I want to compute the limit without using that! – user92360 Oct 07 '15 at 18:01
2

One way is to write the expression $a(n) = e^{\log a_n}$, then you can use the bonds on $\int_{1}^{n} \log x dx < \sum_{k=1}^{n} \log k < 1+ \int_{1}^{n} \log x dx$ to get the asymptotic result.

Alex
  • 19,262
  • I am looking for a method without using Integral... isn't there any solution using more elementary tools?? – user92360 Oct 07 '15 at 18:03
1

It appears (after looking at several other answers posted here) that it is difficult to avoid the use of Stirling's approximation. Stirling's approximation is given by $$\log n! = n\log\left(\frac{n}{e}\right) + \frac{1}{2}\log (2\pi n) + \frac{1}{12n} + O(n^{-2})$$ and hence $$\log (n!)^{1/n} = \log\left(\frac{n}{e}\right) + \frac{\log (2\pi n)}{2n} + O(n^{-2})$$ and therefore $$(n!)^{1/n} = \exp\left(\log\left(\frac{n}{e}\right) + \frac{\log (2\pi n)}{2n} + O(n^{-2})\right) = \frac{n}{e}\exp\left(\frac{\log (2\pi n)}{2n} + O(n^{-2})\right)$$ and thus we get $$(n!)^{1/n} = \frac{n}{e}\left(1 + \frac{\log (2\pi n)}{2n} + O(n^{-2}(\log n)^{2})\right)$$ or $$(n!)^{1/n} = \frac{n}{e} + \frac{\log(2\pi n)}{2e} + O(n^{-1}(\log n)^{2})$$ and then using the above formula for $((n + 1)!)^{1/(n + 1)}$ and on subtracting we get $$\sqrt[n + 1]{(n + 1)!} - \sqrt[n]{n!} = \frac{1}{e} + \frac{1}{2e}\log\left(\frac{n + 1}{n}\right) + O(n^{-1}(\log n)^{2}) $$ and the result follows by taking limit as $n \to \infty$.

I later checked the linked question and found that the result in that question crucially depends on the limit in this question and hence the problem in current question is the more fundamental one and its genuine answer can not be based on the answer of the linked question. Also the linked question has an excellent answer by Daniel Fischer which uses the Stirling's approximation to obtain a limit which is same as the one in current question.

Community
  • 1
Paramanand Singh
  • 87,309
  • Why was this downvoted? – marty cohen Aug 04 '16 at 05:17
  • @martycohen: thanks for the concern. Nowadays it has become fashionable to vote without any comment and i have been the victim many times. If you can see other (deleted) answers of this question, I have downvoted with comment and the people have then deleted their post. That is the way downvotes help. I wonder what's the benefit of downvoting without comment. – Paramanand Singh Aug 04 '16 at 06:53
  • @ParamanandSingh (+1) I think this is the only helpful answer in this posting. – Sungjin Kim Jul 12 '21 at 20:32
0

By Stolz-Cesaro theorem , we have

$$\lim_{n\to\infty}\frac{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}}1=\lim_{n\to\infty}\frac{\sqrt[n]{n!}}n=\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}$$

and by the root-to-ratio limit, we also have

$$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\times\frac{n^n}{n!}=\lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^n}$$

which, by the limit definition of $e$, is given by

$$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\frac1e$$

THIRUMAL 5688
  • 534