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This is about a homework I have to do. I don't want the straight answer, just the hint that may help me start on this.

To give you context, we worked on series, and we're now studying integrals, linking the two with Riemann sums.

Now here is the question :

Using Riemann sum, prove :

$$ \lim_{n\to\infty} \frac{1}{n} \sqrt[n]{n!} = 1 $$

Which is to say, find $a,b\in \mathbb{R}$, $f$ a function, so that :

$$ \frac{1}{n} \sqrt[n]{n!} = \left(b-a\over n\right)\sum_{k=1}^n f\left(a+k{b-a\over n}\right) $$

Any help would be greatly appreciated, thanks.

EDIT : Thanks to the answers and after looking at it for a while, it appears the limit is $1\over e$ instead of $1$

Another User
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Furrane
  • 1,562

2 Answers2

6

$$ A=\frac{1}{n} \sqrt[n]{n!} \\= \sqrt[n]{\frac{n!}{n^n}}\\\sqrt[n]{\frac{n(n-1)(n-2)...3.2.1}{n.n.n.n.n...n}}\\$$ now take log $$\ln A=\ln \sqrt[n]{\frac{n(n-1)(n-2)...3.2.1}{n.n.n.n.n...n}}\\=\frac{1}{n}\ln(\frac{n(n-1)(n-2)...3.2.1}{n.n.n.n.n...n})=\\\frac{1}{n}\ln(\frac{1.2.3...(n-3)(n-2)(n-1)n}{n.n.n.n.n...n})=\\ \frac{1}{n}(\ln\frac{1}{n}+\ln\frac{2}{n}+\ln\frac{3}{n}+...+\ln\frac{n}{n})=\\\frac{1}{n} \Sigma_{i=1}^{n}\ln(\frac{i}{n})$$ and now $$\lim \ln A=\int_{0}^{1}\ln(x)dx=x\ ln x-x =0-1-0-0\\ \ln A =-1 \to A=\frac{1}{e}$$

Khosrotash
  • 24,922
4

Hint:

your sequence is

$a_n=(\frac{n!}{n^n})^{ \frac{1}{n} }$

take logarithm

$S_n=ln(a_n)=\frac{1}{n}\sum_{k=1}^nln(\frac{k}{n})$

$\lim_{n\to \infty}S_n=\int_0^1ln(x)dx$

which gives $-1$.

it seems that you sequence goes to

$\frac{1}{e}$.

hamam_Abdallah
  • 62,951