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Prove that the $\lim_{n \rightarrow \infty} \frac{2^{n} n!}{n^{n}} = 0$

$\rightarrow \frac{2^{n} n!}{n^{n}} = $ $(\frac{2}{n})^{n} n!$

Its possible to say that $\lim_{n \rightarrow \infty} $$\frac{2}{n}$ is $0$ and because of this reason $\lim_{n \rightarrow \infty} $$(\frac{2}{n})^{n}$ is $0$ also ? And Because of this $\lim_{n \rightarrow \infty} \frac{2^{n} n!}{n^{n}} = 0$ ?

Thanks.

David K
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NM2
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4 Answers4

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Let $a_n=\dfrac{2^{n} n!}{n^{n}}$. Then $$ \dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1} (n+1)!}{(n+1)^{n+1}} \dfrac{n^{n}}{2^{n} n!} = 2 \left(\dfrac{n}{n+1}\right)^n = 2 \dfrac{1}{\left(1+\dfrac{1}{n}\right)^n} \to \dfrac2e < 1 $$

The ratio test implies that the series $\sum a_n$ converges and so $a_n \to 0$.

You can avoid the ratio test by noting that $\dfrac2e<0.75$ and so $\dfrac{a_{n+1}}{a_n} < 0.75$ for $n \ge N$, which gives $a_n < 0.75^{n-N} a_N \to 0$.

lhf
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  • Definitely missed this trick. +1 – Chinny84 Dec 11 '15 at 15:21
  • Hey, You just prooved that $a_{n}$ is monotonic decreasing, but how do you it convergeas to $0$ ? Thanks. – NM2 Dec 12 '15 at 10:03
  • @Noam, if a series converges then the sequence of its terms converges to $0$. – lhf Dec 12 '15 at 13:45
  • But how do you know it converges ? For example, $a_{1} = 1 $ and $a_{n+1} = a_{n} -1 $ is decreasing but not converging. – NM2 Dec 12 '15 at 16:33
  • OK. The ratio test is less than 1. So the series converges. But how do you it converges to $0$ ? Thanks. – NM2 Dec 13 '15 at 07:46
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Not sure what you can use, so try $t= e^{\log t}$ and then notice that $\log n! = \sum_{k=1}^{n} \log k \sim n \log -n +1$, compared to the integral. Then you'll be able to cancel out a few terms. Can you handle from here?

Alex
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Bernoulli's Inequality says that $\left(1+\frac1n\right)^n$ is an increasing sequence. Thus, for $n\ge2$, $$ \begin{align} \frac{a_{n+1}}{a_n} &=\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\\[3pt] &=\frac2{\left(1+\frac1n\right)^n}\\ &\le\frac2{\left(\frac32\right)^2}\\ &=\frac89 \end{align} $$ Therefore, for $n\ge2$, $$ a_n\le a_2\left(\frac89\right)^{n-2} $$ Thus, we can say that for $n\ge2$, $$ 0\le\frac{2^nn!}{n^n}\le2\left(\frac89\right)^{n-2} $$ and by the Squeeze Theorem, we have $$ \lim_{n\to\infty}\frac{2^nn!}{n^n}=0 $$

robjohn
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  • How to apply Bernoulli's Inequality to show $(1+\frac{1}{n})^n$ is an increasing sequence? I can only see that from computing $(1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n$ – robit Dec 13 '15 at 14:31
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    Bernoulli says $$\left(1+\frac1{n+1}\right)^{\frac{n+1}n}\ge1+\frac{n+1}n\frac1{n+1}$$ raise both to the power $n$ $$\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n$$ – robjohn Dec 13 '15 at 14:54
  • It's a neat trick :D – robit Dec 13 '15 at 15:09
  • @robit: It's not quite Bernoulli's inequality, because robjohn is using for real exponent, which isn't Bernoulli's and cannot be proven by induction. The easiest way to prove the real exponent version is via asymptotic expansion, which makes the inequality redundant as you saw in my other answer. However, there is a way to use the integer exponent version to prove the monotonicity.. $\dfrac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n} = ( 1 - \frac{1}{(n+1)^2} )^{n+1} \frac{n+1}{n} \ge ( 1 - \frac{1}{n+1} ) \frac{n+1}{n} = 1$ for any integer $n > 0$. – user21820 Dec 15 '15 at 05:09
  • @user21820: If you check, Bernoulli's Inequality can be generalized, by induction, for rational exponents as well. Then, by continuity, for real exponents. See this answer for a proof of the integer case and this answer for the rational case. – robjohn Dec 15 '15 at 06:10
  • @robjohn: I know it can be generalized, just that in the general case it's far easier to prove it using asymptotic expansion. I'm just saying that the trick I showed can be used to obtain the desired inequality using just the integer version, which is the one originally found by Bernoulli. My point is simply that asymptotic expansion is a more powerful tool that can be mechanically applied, while Bernoulli's inequality is a simpler tool. My comment is that one doesn't even require the general version. – user21820 Dec 15 '15 at 06:54
  • @user21820: the inequality you cite above is a special case of the one in the second answer I cited. Using that inequality and induction, we get the full rational version. Once we have that, it makes sense to use it. – robjohn Dec 15 '15 at 23:45
  • If it makes sense to use existing tools, then likewise it's better to use asymptotic expansion since it's superior to Bernoulli's inequality and does not require any thinking. If we want to stick to extremely simple-to-prove tools, then we should similarly stick to the simplest proofs, which is why I proposed the simple one-line proof using the simplest version of Bernoulli's inequality. In other words, I've nothing against proving and then using more powerful tools, but then in this case we might as well go all the way to asymptotic expansion. That's all. – user21820 Dec 16 '15 at 05:04
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  • noted: $x=2^nn!$ and $y=n^n$

1- simplification:

$\frac{ln(x)}{ln(y)}=\frac{nln(2)}{nln(n)}+\frac{ln(n!)}{ln(n^n)}$

2- property:

$\lim \frac{x}{y}=0$ means $\lim (\ln \frac{x}{y})=ln(0)=-\infty$

$lim(ln(y)-ln(x))=\infty$ $\rightarrow$ $y>x$

from the graph shown if $y>x$ then $y-ln(y)>x-ln(x)$ which means $y-x>ln(y)-ln(x)$

$lim(lny-lnx)<lim(y-x)$ $\rightarrow$ $lim(y-x)=\infty$ $\rightarrow$ $lim(lne^y-lne^x)=\infty$

$\rightarrow$ $lim (ln\frac{e^y}{e^x})=\infty$ $\rightarrow$ $lim (ln\frac{e^x}{e^y})=-\infty$ $\rightarrow$ $$lim\frac{e^x}{e^y}=0$$

3- Calculate the limit

$lim_{n->\infty} \frac{ln(2)}{ln(n)}+\frac{ln(n!)}{ln(n^n)}=0+0=0$

$lim_{n->\infty} \frac{ln(x)}{ln(y)} =0$ means

$lim_{n->\infty} \frac{e^{ln(x)}}{e^{ln(y)}} = 0$

$$lim_{n->\infty} \frac{x}{y} = 0$$

Abr001am
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