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

[juantheron]

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]

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]

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

$\chi$ $\sigma$

 

[juantheron]

Thanks chisigma

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

 

[chisigma]

Thanks chisigma

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

It consists in the Stirling's approximation that in more precise form is...

$\displaystyle \ln n! = n\ \ln n - n + \mathcal {O} (\ln n)\ (1)$ Stirling's Approximation -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

 

[Evgeny.Makarov]

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]

Thanks chisigma

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

Alternatively... in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html#post2494

...it has been demonstrated that for x 'large enough' is... $\displaystyle \phi(x) = \frac{d}{d x} \ln x! \sim \ln x\ (1)$... so that... $\displaystyle \ln n! \sim \int_{0}^{n} \ln x\ dx = n\ \ln n - n\ (2)$ Kind regards$\chi$ $\sigma$