1. ## Correction

Hi guys
Can anyone put me right on this? I keep coming up with the opposite answer.

$\displaystyle a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$
$\displaystyle (n\geq{1})$
Show that $\displaystyle a_{n}$ is monotonic increasing.

Anyway I do it, I keep finding $\displaystyle a_{n}$ is monotonic decreasing. I've tried doing it as $\displaystyle a_{n+1}-a_{n}\geq{0}$, and I've gone very simple and just substituted different random values in for $\displaystyle a_{n}$ to see what kind of resuts it would yield, but it always ends up as $\displaystyle a_{n+1}\leq{a_{n}}$. Can anyone show me where I'm going wrong please??

Thanks,
Jo

2. Originally Posted by simplysparklers
Hi guys
Can anyone put me right on this? I keep coming up with the opposite answer.

$\displaystyle a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$
$\displaystyle (n\geq{1})$
Show that $\displaystyle a_{n}$ is monotonic increasing.

Anyway I do it, I keep finding $\displaystyle a_{n}$ is monotonic decreasing. I've tried doing it as $\displaystyle a_{n+1}-a_{n}\geq{0}$, and I've gone very simple and just substituted different random values in for $\displaystyle a_{n}$ to see what kind of resuts it would yield, but it always ends up as $\displaystyle a_{n+1}\leq{a_{n}}$. Can anyone show me where I'm going wrong please??

Thanks,
Jo
$\displaystyle a_{n+1} - a_{n} = \frac{3a_n + 1}{a_n + 3} - a_n = \frac{1 - a_n ^2}{3 + a_n}$

In the previous thread we proved that $\displaystyle 0 \leq a_n < 1$.This implies $\displaystyle 1 - a_n ^2 > 0, 3+a_n > 0$. These two together imply $\displaystyle a_{n+1} - a_{n} > 0$

3. Originally Posted by simplysparklers
Hi guys
Can anyone put me right on this? I keep coming up with the opposite answer.

$\displaystyle a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$
$\displaystyle (n\geq{1})$
Show that $\displaystyle a_{n}$ is monotonic increasing.

...
1.
$\displaystyle a_{n+1}-a_n=\frac{3 \cdot \frac{3a_n+1}{a_n+3}+1}{\frac{3a_n+1}{a_n+3}+3}- \frac{3a_n+1}{a_n+3} = \frac{6a_n+13-3(a_n)^2}{(a_n+3)(3a_n+5)}$

Since $\displaystyle a_n<1$ for all n this fraction is positiv thus the sequence is monotonic increasing

2. I've attached a screenshot of a spreadsheet calculating $\displaystyle a_n$

3. What are you doing in Bamberg?

4. Thanks Earboth!!
& I'm on my Erasmus year in Bamberg!I study German and Maths, and I have maths work sent over from my home uni to be done!Awful book to work from, and no lecturer to help me, that's why I'm so grateful I found you MHF guys!!