Let $ x_{1} = 8 $ and $ x_{n+1} = \frac{1}{2} x_n + 2 $ for n = 1, 2, … . Show {$x_n$} is bounded and monotone, and find its limit.
I know to show that this sequence is bounded, I need to find a convergent subsequence. I see that at n = 4, the sequence is $ 4\frac{1}{2}, 4\frac{1}{4}, 4\frac{1}{8}, 4\frac{1}{16} $ …, but how do I prove that it is in fact a subsequence that can be written as {$x_n$} = $ 4 + \frac{1}{2^n} $
I do not know how to prove that the sequence is monotonic.
I know that the limit is 4, but I am unsure how to prove it.
Best Answer
The lower bound of 4 can be shown by a quick induction: if $x_n>4$ then $x_{n+1}=\frac{x_n}{2}+2>\frac{4}{2}+2=4$, and since $x_1=8>4$ we have $x_n>4$ for all $n$.
Now that we have a lower bound, we can also easily show that $\{x_n\}$ is monotone decreasing: $$x_{n+1}-x_n=\frac{1}{2}x_n+2-x_n=2-\frac{x_n}{2}<2-\frac{4}{2}=0,$$ thus $x_{n+1}<x_n$.