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?
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