Let x(1) := 8 and x(n+1) := $\frac12$x$_{n}$ + 2. Show that (x$_n$) is bounded and monotone. Find the limit. I have shown that (x$_n$) is decreasing and bounded below by 4 using induction and I know that the lim(x$_n$) should = inf{x$_n$ : n $\in$ $\mathbb N$} but I am unsure of how to find this value.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ I assume you mean to say $x_{n + 1} = \frac 1 2 x_n + 2$. $\endgroup$– user61527Jun 15, 2014 at 20:06
-
$\begingroup$ If the sequence is bounded below by $4$ and is decreasing, it has a limit of $4$. Any sub-sequence of $x_{n}$ have the same limit as $x_{n}$. Because, if there was two different sub-sequences of $x_{n}$ with different limits, then $x_{n}$ wouldn't have limit. $\endgroup$– user24047Jun 15, 2014 at 22:11
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
If $L$ is the limit, then it had better satisfy
$$L = \frac 12 L + 2$$
Do you see why? One way to verify this is to prove that any larger number can't be the limit of the sequence - for the next term in the sequence would have to be quite a bit smaller.