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Proving (n/2)^n > n! > (n/3)^n for all n > 6

jacks

Well-known member
Apr 5, 2012
226
If $A = (300)^{600}$ and $B=600!$ and $C = (200)^{600}$, Then arrangement of $A,B,C$ in Decreasing order, is

My approach:: Here $A>C$ and Using the formula $\displaystyle \left(\frac{n}{2}\right)^n>n! >\left(\frac{n}{3}\right)^n$ for all natural no. $n>6$

and put $n=600,$ we get $(300)^{600}>600!>(200)^{600}$ So $A>B>C$

My Question is How can we prove $\displaystyle \left(\frac{n}{2}\right)^n>n! >\left(\frac{n}{3}\right)^n$ for all natural no. $n>6$

Can we prove it without induction.

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
If $A = (300)^{600}$ and $B=600!$ and $C = (200)^{600}$, Then arrangement of $A,B,C$ in Decreasing order, is

My approach:: Here $A>C$ and Using the formula $\displaystyle \left(\frac{n}{2}\right)^n>n! >\left(\frac{n}{3}\right)^n$ for all natural no. $n>6$

and put $n=600,$ we get $(300)^{600}>600!>(200)^{600}$ So $A>B>C$

My Question is How can we prove $\displaystyle \left(\frac{n}{2}\right)^n>n! >\left(\frac{n}{3}\right)^n$ for all natural no. $n>6$

Can we prove it without induction.

Thanks
This should follow from Stirling's formula together with the fact that $2<e<3$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
If $A = (300)^{600}$ and $B=600!$ and $C = (200)^{600}$, Then arrangement of $A,B,C$ in Decreasing order, is

My approach:: Here $A>C$ and Using the formula $\displaystyle \left(\frac{n}{2}\right)^n>n! >\left(\frac{n}{3}\right)^n$ for all natural no. $n>6$

and put $n=600,$ we get $(300)^{600}>600!>(200)^{600}$ So $A>B>C$

My Question is How can we prove $\displaystyle \left(\frac{n}{2}\right)^n>n! >\left(\frac{n}{3}\right)^n$ for all natural no. $n>6$

Can we prove it without induction.

Thanks
The proof is made easier if You consider logarithms, because...

$\displaystyle \ln\ (\frac{n}{2})^{n} = n\ (\ln n - \ln 2)\ (1)$

$\displaystyle \ln\ (\frac{n}{3})^{n} = n\ (\ln n - \ln 3)\ (2)$

$\displaystyle \ln n! \sim n\ (\ln n - 1)\ (3)$

Take into account that $\displaystyle \ln 2 < 1 < \ln 3$...

Kind regards

$\chi$ $\sigma$
 

jacks

Well-known member
Apr 5, 2012
226
Thanks chisigma

would you like to explain me How we get $\ln(n!)\approx n\left(\ln (n)-1\right)$
 

chisigma

Well-known member
Feb 13, 2012
1,704

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
The lower bound $n!>\left(\frac{n}{3}\right)^n$ is easy to obtain from the Maclaurin series for $e^x$. We have $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$. In particular, $e^x>\frac{x^n}{n!}$. Taking $x=n$ gives $e^n>\frac{n^n}{n!}$, i.e., $n!>\left(\frac{n}{e}\right)^n> \left(\frac{n}{3}\right)^n$ since $e<3$.
 

chisigma

Well-known member
Feb 13, 2012
1,704