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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.

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  • $\begingroup$ can you use Stirling approximation? $\endgroup$
    – Chinny84
    Dec 11, 2015 at 14:54
  • $\begingroup$ No, I can't use it. $\endgroup$
    – NM2
    Dec 11, 2015 at 14:56
  • $\begingroup$ Ok, so I have this $0< \frac{k}{n}\leq 1$ .. but what's next ? $\endgroup$
    – NM2
    Dec 11, 2015 at 15:11

4 Answers 4

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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 $a_n < 0.75^{n-N} a_N \to 0$.

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  • $\begingroup$ Definitely missed this trick. +1 $\endgroup$
    – Chinny84
    Dec 11, 2015 at 15:21
  • $\begingroup$ Hey, You just prooved that $a_{n}$ is monotonic decreasing, but how do you it convergeas to $0$ ? Thanks. $\endgroup$
    – NM2
    Dec 12, 2015 at 10:03
  • $\begingroup$ @Noam, if a series converges then the sequence of its terms converges to $0$. $\endgroup$
    – lhf
    Dec 12, 2015 at 13:45
  • $\begingroup$ But how do you know it converges ? For example, $a_{1} = 1 $ and $a_{n+1} = a_{n} -1 $ is decreasing but not converging. $\endgroup$
    – NM2
    Dec 12, 2015 at 16:33
  • $\begingroup$ OK. The ratio test is less than 1. So the series converges. But how do you it converges to $0$ ? Thanks. $\endgroup$
    – NM2
    Dec 13, 2015 at 7:46
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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?

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

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  • $\begingroup$ How to apply Bernoulli's Inequality to show $(1+\frac{1}{n})^n$ is an increasing sequence? I can only see that from computing $(1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n$ $\endgroup$
    – robit
    Dec 13, 2015 at 14:31
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    $\begingroup$ 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 $n$ $$\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n$$ $\endgroup$
    – robjohn
    Dec 13, 2015 at 14:54
  • $\begingroup$ It's a neat trick :D $\endgroup$
    – robit
    Dec 13, 2015 at 15:09
  • $\begingroup$ @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 $n > 0$. $\endgroup$
    – user21820
    Dec 15, 2015 at 5:09
  • $\begingroup$ @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. $\endgroup$
    – robjohn
    Dec 15, 2015 at 6:10
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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

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

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

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