Limit involving factorial

Question

Can someone give me a solution to this problem, I'm lost.

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

Answer 1

Following LStU's hint in the comments, we observe first that

$$ n! \sim \left(\frac ne \right)^n \sqrt{2\pi n} $$

so

$$ (n!)^{1/n} \sim \frac ne \sqrt[2n]{2\pi n} $$

so

$$ \frac{(n!)^{1/n}}{n} \sim \frac1e \sqrt[2n]{2\pi n} $$

Keeping in mind that $n^{1/n} \to 1$, can you proceed to the desired limit?

Answer 2

Hint: Use the theorem that if $a_{n} >0$ and $a_{n+1}/a_{n}\to L$ then $a_{n} ^{1/n}\to L$. Take $a_n=n! /n^{n} $. You should easily get the answer as $1/e$.

Answer 3

I propose here a way that doesn't require Stirling:

Lemma: We have that $\ln(n!) = \sum_{i=2}^n \ln(n)$. Since $\ln(\cdot )$ is strincly increasing and $\ln(x) \geq 0, \forall x \geq 1$ we have $$ \int_1^n \ln(x) \leq \sum_{i=2}^n \ln(n) \leq \int_2^{n+1} \ln(x) \leq \int_1^{n+1} \ln(x) \implies\\ n\ln(n)-n+1 \leq \ln(n!) \leq (n+1)\ln(n+1)-n $$

Now getting back to the limit. We have that $\ln\left( \frac{(n!)^{1/n}}{n}\right)$ = $\frac{\ln(n!)}{n} - \ln(n) $ can be bounded using the Lemma: $$ \frac{1}{n} - 1 \leq \ln\left( \frac{(n!)^{1/n}}{n}\right) \leq \frac{\ln(n+1)}{n} + \ln(n+1) -1 - \ln(n) = \frac{\ln(n+1)}{n} -1 + \ln(1 + \frac{1}{n}) \leq \frac{\ln(2n)}{n} + \ln(1+\frac{1}{n}) - 1 = \frac{\ln(2)}{n} + \frac{\ln(n)}{n} + \ln(1 + \frac{1}{n}) -1 $$

Now taking the limit and knowing that $\lim_{n\to \infty} \ln(n)/n = 0$ and then taking back the exponential should give you the result.

Answer 4

In the same spirit as in answers and comments, consider $$A_n=\frac{(n!)^{\frac{1}{n}}}{n}\implies \log(A_n)=\frac{\log (n!)}{n}-\log (n)$$ and use Stirling formula for large $n$ $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+O\left(\frac{1}{n}\right)$$ which makes $$\log(A_n)=-1+\frac{\log (2 \pi )+\log \left({n}\right)}{2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

In fact, even for small values of $n$, this approximation leads to quite accurate values $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 0.922137 & 1.000000 \\ 2 & 0.692641 & 0.707107 \\ 3 & 0.600144 & 0.605707 \\ 4 & 0.550472 & 0.553341 \\ 5 & 0.519303 & 0.521034 \\ 6 & 0.497813 & 0.498966 \\ 7 & 0.482039 & 0.482859 \\ 8 & 0.469932 & 0.470544 \\ 9 & 0.460323 & 0.460796 \\ 10 & 0.452496 & 0.452873 \end{array} \right)$$