# Thread: Prove a sequence is monotone and bounded, and find the limit.

1. ## Prove a sequence is monotone and bounded, and find the limit.

One last thing. So I have this sequence:

$\displaystyle x_1:= 8 \ \mbox{and} \ x_{n+1} =\frac{1}{2}x_n + 2 \ \ \mbox{for all} \ n\in N$

I know that it is strictly decreasing, which makes it monotone, but how would I prove that?

I also think that the limit is 4 just by looking at the first few terms, which means that since it converges that it is bounded, but I don't know the proper way to prove that either. Help?

Thanks so much.

2. Originally Posted by mremwo
$\displaystyle x_1:= 8 \ \mbox{and} \ x_{n+1} =\frac{1}{2}x_n + 2 \ \ \mbox{for all} \ n\in N$ I know that it is strictly decreasing, which makes it monotone, but how would I prove that?

Use induction.

I also think that the limit is 4 just by looking at the first few terms, which means that since it converges that it is bounded, but I don't know the proper way to prove that either.

Take limits in both sides of $\displaystyle x_{n+1}=x_n/2+2$ .

P.S. You need also prove that the sequence is lower bounded (choose for example $\displaystyle 0$ as a lower bound) .

Fernando Revilla

3. Originally Posted by mremwo
One last thing. So I have this sequence:

$\displaystyle x_1:= 8 \ \mbox{and} \ x_{n+1} =\frac{1}{2}x_n + 2 \ \ \mbox{for all} \ n\in N$

I know that it is strictly decreasing, which makes it monotone, but how would I prove that?

I also think that the limit is 4 just by looking at the first few terms, which means that since it converges that it is bounded, but I don't know the proper way to prove that either. Help?

Thanks so much.
The difference equation can be written as...

$\displaystyle \displaystyle \Delta_{n}= x_{n+1} - x_{n} = 2 - \frac{x_{n}}{2} = f(x_{n})$ (1)

The f(*) is linear and is represented here [black line]...

There is only one 'attractive fixed point' at $\displaystyle x_{0}=4$ and because for all x is $\displaystyle |f(x)|\le |x_{0}-x|$ , all the possible $\displaystyle x_{1}$ will produce a sequence monotonically convergent in $\displaystyle x_{0}$. Of course if $\displaystyle x_{1}< x_{0}$ the sequence will be increasing, if $\displaystyle x_{1}>x_{0}$ decreasing...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. To show that it is monotone :
$\displaystyle x_1=8$ and $\displaystyle x_2=6$ so $\displaystyle x_1>x_2$ . Suppose that $\displaystyle x_n>x_{n+1}$ then : $\displaystyle \frac{1}{2}x_n>\frac{1}{2}x_{n+1}\Rightarrow\frac{ 1}{2}x_n+2>\frac{1}{2}x_{n+1}+2 \Rightarrow x_{n+1}>x_{n+2}$ so it's decreasing.