I'm working on Exercise $3.3.2$ from Introduction to Real Analysis by Bartle and Sherbert, which says:
Let $x_1>1$ and $x_{n+1}:=2-1/x_n$ for $n\in\mathbb{N}.$ Show that $(x_n)$ is bounded and monotone. Find the limit.
First, to get an idea of the behavior of the sequence to see whether it was increasing or decreasing, I took as an example $x_1=2$ and deduced that $x_2=1\frac{1}{2},x_3=1\frac{1}{3},x_4=1\frac{1}{4},$ etc., so it seems the sequence must be decreasing.
So now I'm trying to prove that $(x_n)$ is decreasing by proving the statement $P(n):x_{n+1}\leq x_n$ is true for all $n\in\mathbb{N}$ and I got stuck at the base case, $P(1):x_2\leq x_1$. If at all possible, I'd only like help with the base case of just the proof of monotonicity so that I can use that support as a springboard to figure out the rest of the problem on my own.
I deduced so far that since $x_1>1,-1/x_1>-1.$ (right?) So then
$$ x_2=2-\frac{1}{x_1}>2-1=1.$$
That only tells me that $x_2>1,$ and I don't think this really helps me show that $x_2\leq x_1.$ (However, this will help me prove that $(x_n)$ is bounded when that time comes…)
I also tried:
$$x_2=2-\frac{1}{x_1}=\frac{2x_1-1}{x_1}<2x_1-1<2x_1$$
But surely $x_2<2x_1$ doesn't imply that $x_2\leq x_1$…
Are there any other ways to show $x_2\leq x_1$?
To be clear, please only give me a hint for the base case above — I want to try to figure out the rest on my own. Thank you in advance for your help! 🙂
Best Answer
By induction $x_n>1$ for all $n$, so $$0<( {x_n}^{1/2}- x_n^{-1/2})^2=x_n-2+1/{x_n} =x_n-x_{n+1}\,.$$