-1

I am looking for $$\lim \limits_{n \to \infty}{2^nn! \over n^n}$$ I check by inspection that the limit is $0$. My first suspicion is that we might be able to use the following fact:

$${a_{n+1} \over a_n} \to g \implies \sqrt[n]{a_n} \to g$$

I rewrite $${2^nn! \over n^n} = \sqrt[n]{\left({2^nn! \over n^n}\right)^n}$$ set $$a_n=\left({2^nn! \over n^n}\right)^n$$ and compute $${a_{n+1} \over a_n}={2^{2n+1}n! \over (n+1)^n}\left({1 \over 1 + {1 \over n}} \right)^{n^2}$$ The second term seems to be a natural candidate for a limit with $e$, however, I can't figure out the limit for the first term.

Alternatively, there might be an obvious candidate for the squeeze theorem for the original limit which I'm missing.

Did
  • 279,727
Zelazny
  • 2,489
  • 1
    Your approach is odd. Why not consider directly $$\lim \limits_{n \to \infty}{x_{n+1} \over x_n}$$ with $$x_n={2^nn! \over n^n}\ ?$$ – Did Nov 16 '16 at 20:28
  • The oddity stems from the fact that the material I'm working with hasn't discussed the ratio test in the form you're describing at all, only the form with the $n$th root. – Zelazny Nov 16 '16 at 21:36

2 Answers2

4

Notice the series

$$ \sum \frac{ 2^n n! }{n^n} $$ converges by the ratio test since

$$ \frac{ a_{n+1} }{a_n} = \frac{2^{n+1} (n+1)!}{(n+1)^{n+1} } \frac{n^n}{n! 2^n} = 2 \left( \frac{n}{n+1} \right)^n \to \frac{2}{e} <1 $$

Thus,

$$\boxed{ \lim_{n \to \infty}\frac{ 2^n n! }{n^n} = 0 }$$

ILoveMath
  • 10,694
  • 5
    If I may ask : Why introduce the series ? The simple fact that ${a_{n+1}\over a_n}\to \alpha \lt 1$ proves that $a_n$ is dominated by $\beta^n$ for some $\alpha \lt \beta \lt 1$ and thus converges to 0, no ? – Astyx Nov 16 '16 at 20:37
  • @Astyx I think this is a negative quirk of the standard calculus pedagogy. People are taught to reach for the ratio test in the context of series, and only in the context of series. Yet really it is just is just a systematic way to find a number $a$ such that your given sequence either dominates or is dominated by $a^n$. As it happens that's convenient in the setting of series, because given such a number one can then apply the limit comparison test, but it doesn't really have anything intrinsically to do with series. – Ian Nov 16 '16 at 21:15
3

For very large $n$ you can use Stirling approximation for the factorial:

$$n! \approx \sqrt{2\pi n} \frac{n^n}{e^n}$$

hence your limit becomes

$$\lim_{n\to +\infty} \frac{2^n}{n^n}\sqrt{2\pi n} \frac{n^n}{e^n}$$

$$\lim_{n\to +\infty} 2^n \frac{\sqrt{2\pi n}}{e^n}$$

$$\sqrt{2\pi} \lim_{n\to +\infty} \frac{2^n n^{1/2}}{e^n}$$

We can solve this sending all into the exponential representation

$$2^n = e^{n\ln (2)}$$

$$n^{1/2} = e^{1/2\ ln(n)}$$

So

$$\sqrt{2\pi} \lim_{n\to +\infty} \frac{e^{n\ln (2) + 1/2\ \ln(n)}}{e^n}$$

Unifying all the exp

$$\sqrt{2\pi} \lim_{n\to +\infty} e^{n\ln (2) + 1/2\ \ln(n) - n}$$

$$\sqrt{2\pi} \lim_{n\to +\infty} e^{n[\ln(2) - 1] + 1/2 \ln(n)}$$

The quantity $\ln(2) - 1$ is negative, and we will call it $-\alpha$. Numerically it is $-0.3068...$

Hence

$$\sqrt{2\pi} \lim_{n\to +\infty} e^{-\alpha n + 1/2 \ln(n)}$$

$n$ wins over $\ln(n)$ so the limit is basically

$$\lim_{n\to +\infty} e^{-\alpha n} = \boxed{0}$$

Enrico M.
  • 26,114
  • Using the Stirling approximation and then giving so much detail as to why $\left({2\over e }\right)^n \sqrt{n}$ seems weird, but this works. – Astyx Nov 16 '16 at 21:01
  • 1
    @Astyx Whenever I see a factorial and a limit at infinity, the first thought is Stirling because.. it works perfectly :D The other "approximations" are not that bad, but only because $n$ is very very large. Maybe it's a bit weird but it work good to me. Hope it's not too "broad". – Enrico M. Nov 16 '16 at 21:03
  • Yes, what I meant is that if one knows the Stirling equivalent, they would probably know that geometric series are preponderant in comparison to powers of n, and since $2\over e\lt1$, the sequence goes to 0 ... but maybe I'm wrong :) – Astyx Nov 16 '16 at 21:08