0
$\begingroup$

As I am trying to calculate the lim of $(n+1)!/n^n$ when n tends to infinity,

Can I say that it is equivalent to $(n+1)*n!/n^n$ and since $n!/n^n$ = $exp(-n)$ tends to 0.

Then $lim (n+1)!/n^n$ = $lim (n+1) * $exp(-n) which is like doing $lim (n+1) * 0 $ when n -> infinity?

Or I can't do this trick ?

  • 0
    No you cannot. A quantity of the form $0\cdot \infty$ is called *indeterminate*. – 2017-01-20
  • 2
    Can you say that $n\cdot\frac{1}{n}\to 0$ because $\frac{1}{n}\to 0$? – 2017-01-20
  • 0
    Why not ? since lim (n!)/n^n = lim exp(-n) --> 0 then (n+1) n! / n^n = (n+1) * exp(-n) = n/exp(n) + 1/exp(n) = 0 + 0 = 0 .. – 2017-01-20
  • 1
    Also, the quantity you have in the question header is different from what you have typed in the body of the question. Which quantity is it? – 2017-01-20
  • 0
    Edited the header, gonna fix the body – 2017-01-20

2 Answers 2

3

The proper way to do this is to use Stirling's Approximation which says that for large $n$

$$ n! \;\; \approx \;\; n^n e^{-n}\sqrt{2\pi n}. $$

Therefore in your computation you should have that

$$ \frac{(n+1)!}{n^n} \;\; =\;\; (n+1) \frac{n!}{n^n} \;\; \approx \;\; (n+1)e^{-n} \sqrt{2\pi n} \;\; \sim \;\; \frac{n^{3/2}}{e^n} + \frac{\sqrt{n}}{e^n}. $$

Using L'Hopital's rule, both of these quantities go to zero.

  • 0
    I think it is the same thing, what I did was I said lim n!/n^n = lim exp(ln(n!)) / exp (nln(n)) then I said that ln(n!) ~ nln(n)-n Hence, we get exp(nln(n)-n-nln(n) ) = exp(-n). So when n tends to infinity, exp(-n) tends to 0. So lim (n+1)! / n^n = lim (n+1) n!/n^n = lim (n+1) exp(-n) = n/exp(n)+1/exp(n) = 0 + 0 = 0 when n tends to infinity.. I just wanna know if that's correct now – 2017-01-20
2

Here is an easier solution in case you forget or don't know Stirling's formula.

$$\frac{(n+1)!}{n^n}=\frac{(n+1)!}{(n+1)^{n}}\frac{(n+1)^{n}}{n^n}$$ $$\frac{(n+1)!}{(n+1)^{n}}=\frac{n+1}{n+1}\frac{n}{n+1}\frac{n-1}{n+1}...\frac{2}{n+1}\frac{1}{1}\lt\frac{2}{n+1}\rightarrow0$$ $$\frac{(n+1)^{n}}{n^n}=(1+\frac{1}{n})^n\rightarrow{e}$$ So $$lim_{n\to \infty}\frac{(n+1)!}{n^n}=0$$

  • 1
    Maybe it's easier to note the expression equals $(n+1)(n/n) ((n-1)/n) \cdots \cdot (2/n)(1/n )\le (n+1)\cdot (2/n^2) \to 0.$ – 2017-01-20
  • 0
    @zhw. yeah you are right. it's better. – 2017-01-20