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

**mremwo** · February 2nd 2011, 11:42 PM

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.

**FernandoRevilla** · February 2nd 2011, 11:49 PM

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.

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

**chisigma** · February 3rd 2011, 03:43 AM

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]...

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

$\chi$ $\sigma$

**SENTINEL4** · February 3rd 2011, 04:39 AM

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.

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