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

• Feb 3rd 2011, 12:42 AM
mremwo
Prove a sequence is monotone and bounded, and find the limit.
One last thing. So I have this sequence:

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

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.
• Feb 3rd 2011, 12:49 AM
FernandoRevilla
Quote:

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

Use induction.

Quote:

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 $x_{n+1}=x_n/2+2$ .

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

Fernando Revilla
• Feb 3rd 2011, 04:43 AM
chisigma
Quote:

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

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

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...

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

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

http://digilander.libero.it/luposabatini/MHF104.bmp

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

Kind regards

$\chi$ $\sigma$
• Feb 3rd 2011, 05:39 AM
SENTINEL4
To show that it is monotone :
$x_1=8$ and $x_2=6$ so $x_1>x_2$ . Suppose that $x_n>x_{n+1}$ then : $\frac{1}{2}x_n>\frac{1}{2}x_{n+1}\Rightarrow\frac{ 1}{2}x_n+2>\frac{1}{2}x_{n+1}+2 \Rightarrow x_{n+1}>x_{n+2}$ so it's decreasing.