MHB Which is Larger: $2005!$ or $2^{18000}$?

[anemone]

Which is larger, $2005!$ or $2^{18000}$?

 

[chisigma]

Which is larger, $2005!$ or $2^{18000}$?

[sp]Using the Stirling approximations You have...

$\displaystyle \ln n! \sim (n + \frac{1}{2})\ \ln n - n + \frac{1}{2}\ \ln (2\ \pi)\ (1)$

... so that is $\displaystyle \ln 2005! \sim 13244,536...$

On the other side is...

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

... so that is $\displaystyle \ln 2^{18000} = 12476.649...$ [/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

[sp]Using the Stirling approximations You have...

$\displaystyle \ln n! \sim (n + \frac{1}{2})\ \ln n - n + \frac{1}{2}\ \ln (2\ \pi)\ (1)$

... so that is $\displaystyle \ln 2005! \sim 13244,536...$

On the other side is...

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

... so that is $\displaystyle \ln 2^{18000} = 12476.649...$ [/sp]

Kind regards

$\chi$ $\sigma$

Thank you chisigma for participating in this challenge problem! And thank you for your nice solution using the Stirling approximations route!

Solution proposed by other through induction method:

Spoiler

We show that for all $n$, $n!>\left(\dfrac{n}{3} \right)^n$. This is true for $n=1$.

Suppose it is true for $n$. Then $(n+1)!=(n+1)n!>(n+1)\left(\dfrac{n}{3} \right)$.

Now, $\left(\dfrac{n+1}{3} \right)^{n+1}\left(\dfrac{3}{n} \right)^n=\dfrac{n+1}{3}\left(1+\dfrac{1}{n} \right)^n<\dfrac{(n+1)e}{3}<(n+1)$, this proves that $(n+1)!>\left(\dfrac{n+1}{3} \right)^{n+1}$ which then justifies the claim by induction.

Therefore, $2005!>2004!>668^{2004}>512^{2004}=2^{9\cdot2004}=2^{18036}>2^{18000}$ and we can conclude by now that $2005!$ is the greater of the two.