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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 '15 at 14:54
  • $\begingroup$ No, I can't use it. $\endgroup$ – Noam Dec 11 '15 at 14:56
  • $\begingroup$ Ok, so I have this $0< \frac{k}{n}\leq 1$ .. but what's next ? $\endgroup$ – Noam Dec 11 '15 at 15:11
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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 '15 at 15:21
  • $\begingroup$ Hey, You just prooved that $a_{n}$ is monotonic decreasing, but how do you it convergeas to $0$ ? Thanks. $\endgroup$ – Noam Dec 12 '15 at 10:03
  • $\begingroup$ @Noam, if a series converges then the sequence of its terms converges to $0$. $\endgroup$ – lhf Dec 12 '15 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. $\endgroup$ – Noam Dec 12 '15 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. $\endgroup$ – Noam Dec 13 '15 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 '15 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 '15 at 14:54
  • $\begingroup$ It's a neat trick :D $\endgroup$ – robit Dec 13 '15 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 '15 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 '15 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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