Answer to Question #246577 in Real Analysis for sam

Question

Let x1 > 1 and xn+1 := 2 −

1

xn

for all n ∈ N. Show that {xn} is bounded

and monotone. Find the limit.

Answer 1

Proof. Since x1 > 1, we have that 1/x1 < 1. This means that x2 =

2 − 1/x1 > 1. We’d like to show that xn > 1 for all n. By induction, we

see that if xk > 1, then xk+1 = 2−1/xk > 1 (easy enough?). At this point

we are bounded above by 2 and below by 1. (Note: We could have made

another arguement that it was bounded below by 0. The bound of 1 is a

little better, but the MCT does not require anything more than bounded.)

Now let’s show that it is monotone (decreasing). To do this, we want to

start by looking at

$0 < (x_1 − 1)^2\\ 0 < x^2_1 − 2x_1 + 1\\ 2x_1 < x^2_1 + 1\\ 2 < x_1 + 1/x_1\\ 2 − 1/x_1 < x_1\\ x_2 < x_1$

Ack! That only takes care of the first case. Now we use induction to finish

the monotone.

$x_{k+1} < x_k\\ 1/x_{k+1} > 1/x_k\\ −1/x_{k+1} < −1/x_k\\ 2 − 1/x_{k+1} < 2 − 1/x_k\\ x_{k+2} < x_{k+1}$

To find the limit, we use the same technique as the examples in the book

and class.

$x_{n+1} = 2 − 1/x_n\\ x = 2 − 1/x\\ x = 1$