1. ## Montone Recursive Sequence

A sequence (x_n) is defined recursively as (x_n+1)=(6(x_n)+1)/(2(x_n)+5) with x1=3 for n є N
Prove (x_n) is monotone

By looking at the sequence it is decreasing in a monotone fashion but how do i construct a proof to show this?

2. Originally Posted by i_never_noticed
A sequence (x_n) is defined recursively as (x_n+1)=(6(x_n)+1)/(2(x_n)+5) with x1=3 for n є N
Prove (x_n) is monotone

By looking at the sequence it is decreasing in a monotone fashion but how do i construct a proof to show this?
Perhaps you could try proof by induction.

3. i start that as (x_n+1)<(x_n)<3
6(x_n+1)+1<6(x_n)+1<19
but how do i then divided through is it correct in saying
(6(x_n+1)+1)/(2(x_n+1)+5)<6(x_n)+1/(2(x_n)+5)<19/11??
which would then be X4<X3<X2

4. as 19/11 is X2

5. Originally Posted by i_never_noticed
A sequence (x_n) is defined recursively as (x_n+1)=(6(x_n)+1)/(2(x_n)+5) with x1=3 for n є N
Prove (x_n) is monotone

By looking at the sequence it is decreasing in a monotone fashion but how do i construct a proof to show this?

We form the difference::

$\displaystyle x_{n+1}-x_{n} = \frac{6x_{n} + 1}{2x_{n} +5}- x_{n}$ which after doing some calculations it is equal to:

$\displaystyle -\frac{(2x_{n} +1)(x_{n}-1)}{2x_{n} +5}$ and so we have:

$\displaystyle s_{n+1}-s_{n} =-\frac{(2x_{n} +1)(x_{n}-1)}{2x_{n} +5}$.................................................. .........................................1

Now we see that every term in (1),except $\displaystyle x_{n}-1$ ,is +ve because $\displaystyle x_{n}>0$ for all, n ,which we can prove easily by induction.

So if we can prove $\displaystyle x_{n}>1$ for all n ,then (1) will be -ve and hence the sequence decreasing.

And :$\displaystyle x_{1}=3>1$
next suppose $\displaystyle x_{n}>1\Longrightarrow [(6x_{n}+1>7)\wedge(2x_{n}+5>7)]\Longrightarrow\frac{6x_{n}+1}{2x_{n}+5}=x_{n+1}>1$
SO THE sequence is decreasing