# Prove that it's strictly increasing

• Apr 12th 2008, 04:39 AM
james_bond
Prove that it's strictly increasing
Prove that $\left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right)$ is strictly increasing for $n>1$. Or in other words the recurrence equation is strictly increasing.
• Apr 12th 2008, 05:07 AM
Moo
Hello,

$a_n=\left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right)$

Let's show that $\frac{a_{n+1}}{a_n} > 1$

$a_n=\frac{(n+1)^n}{n^n} \frac{4n+1}{4n}$

$a_{n+1}=\left(1+\frac{1}{n+1} \right)^{n+1}\left(1+\frac 1{4(n+1)}\right)=\frac{(n+2)^{n+1}}{(n+1)^{n+1}} \frac{4n+5}{4(n+1)}$

$\frac{a_{n+1}}{a_n}=\dfrac{(n+2)^{n+1} \overbrace{(4n+5)}^{\text{i \ take \ this \ one}}}{(n+1)^{n+1} 4(n+1)} \dfrac{\overbrace{n^n 4n}^{4n^{n+1}}}{(n+1)^n \underbrace{(4n+1)}_{\text{i \ take \ this \ one}}}$

$\frac{a_{n+1}}{a_n}=\dfrac{4n+5}{4n+1} \dfrac{(n+2)^{n+1}n^{n+1}}{(n+1)^{2n+2}}=\dfrac{4n +5}{4n+1} \dfrac{(n^2+2n)^{n+1}}{(n^2+2n+1)^{n+1}}$

And i have a problem oO
• Apr 12th 2008, 07:40 AM
Mathstud28
If
Quote:

Originally Posted by james_bond
Prove that $\left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right)$ is strictly increasing for $n>1$. Or in other words the recurrence equation is strictly increasing.

you get too encumbered in the $\frac{a_{n+1}}{a_n}$ method you could differntiate and see if it is monotonically increasing on $(1,\infty_$...haha
• Apr 12th 2008, 07:49 AM
Moo
I derivated but it didn't give something interesting ;)
• Apr 12th 2008, 08:59 AM
Mathstud28
Desole
Quote:

Originally Posted by Moo
I derivated but it didn't give something interesting ;)

But what do you mean it didnt give you something interesting?
• Apr 12th 2008, 09:03 AM
Moo
You have to study the sign of the derivative. Well, good luck ^^ I've found it easier to study $\frac{a_{n+1}}{a_n}$ but can't conclude (just too lazy maybe)
• Apr 12th 2008, 09:24 AM
Mathstud28
Haha
Quote:

Originally Posted by Moo
You have to study the sign of the derivative. Well, good luck ^^ I've found it easier to study $\frac{a_{n+1}}{a_n}$ but can't conclude (just too lazy maybe)

You are probably right it is much easier
• Apr 28th 2008, 06:54 PM
JaneBennet
Let $N=\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{4n} \right).$

Then $\ln{N}=n\ln{\left(1+\frac{1}{n}\right)}+\ln{\left( 1+\frac{1}{4n}\right)}$ and N is strictly increasing $\Leftrightarrow$ $\ln{N}$ is strictly increasing.

Let $\mathrm{f}(x)=x\ln{\left(1+\frac{1}{x}\right)}+\ln {\left(1+\frac{1}{4x}\right)}$

Then $\mathrm{f}'(x)=\ldots=\ln{\left(1+\frac{1}{x}\righ t)}+\frac{3}{x+1}-\frac{1}{x}$

This is positive for all $x>1$ because $x>1$ $\Rightarrow$ both $\ln{\left(1+\frac{1}{x}\right)}>0$ and $\frac{3}{x+1}-\frac{1}{x}>0$.

Hence $\ln{N}$ is strictly increasing.