# Math Help - Prove that it's strictly increasing

1. ## 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.

2. 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

3. ## If

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

4. I derivated but it didn't give something interesting

5. ## Desole

Originally Posted by Moo
I derivated but it didn't give something interesting
But what do you mean it didnt give you something interesting?

6. 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)

7. ## Haha

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

8. 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.