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

Asked Jun 03 '21 at 16:09

Active Jun 04 '21 at 05:46

Viewed 77 times

$$(n!)^2=[1\cdot n][2\cdot (n-1)][3\cdot (n-2)] \cdots [(n-1)\cdot 2][n\cdot 1].$$ Each of the $n$ products $(k+1)\cdot (n-k)$, for $0\le k<n$, is $\ge n$.

Thus $$(n!)^2 \ge n^{n} \quad\implies \left(\frac{n!}{(n+1)^n}\right)^{1/n}\ge \frac{\sqrt{n}}{n+1}.$$

Please help to solve it

edited Jun 03 '21 at 18:20

asked Jun 03 '21 at 16:09

user1942348

3,871

4

What do you mean by "find" the expression? – MPW Jun 03 '21 at 16:12
5

Did you just copy https://math.stackexchange.com/q/1319992 ? – Gary Jun 03 '21 at 16:14
For large n use Stirling's formula for n!. – herb steinberg Jun 03 '21 at 17:42
@MPW Sorry ... edited – user1942348 Jun 03 '21 at 18:23
@Gary Yes, Understand and taken help. I wanted to solve it in that style. – user1942348 Jun 03 '21 at 18:24
Take the logarithm. Stirling's formula is not so easy to obtain. But for $n\ge 2$ we have $(n\ln n)-n+1= \int_1^n\ln t dt<\ln (n!)<\int_2^{n+1}\ln t dt=(n+1)\ln (n+1)-(n+1)-(2-\ln 2),$ which is adequate for your Q. – DanielWainfleet Jun 04 '21 at 11:48

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

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