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

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- Apr 12th 2008, 03:39 AMjames_bondProve that it's strictly increasing
Prove that $\displaystyle \left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right)$ is strictly increasing for $\displaystyle n>1$. Or in other words the recurrence equation is strictly increasing.

- Apr 12th 2008, 04:07 AMMoo
Hello,

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

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

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

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

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

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

And i have a problem oO - Apr 12th 2008, 06:40 AMMathstud28If
- Apr 12th 2008, 06:49 AMMoo
I derivated but it didn't give something interesting ;)

- Apr 12th 2008, 07:59 AMMathstud28Desole
- Apr 12th 2008, 08:03 AMMoo
You have to study the sign of the derivative. Well, good luck ^^ I've found it easier to study $\displaystyle \frac{a_{n+1}}{a_n}$ but can't conclude (just too lazy maybe)

- Apr 12th 2008, 08:24 AMMathstud28Haha
- Apr 28th 2008, 05:54 PMJaneBennet
Let $\displaystyle N=\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{4n} \right).$

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

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

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

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

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