Proving Convergence of a Sequence with Bounded and Decreasing Terms

[cragar]

Homework Statement

Prove that the sequence defined by $x_1=3$
and $x_{n+1}= \frac{1}{4-x_n}$

The Attempt at a Solution

Well I found like the first 4 terms of this sequence and it seems to be decreasing, heading closer to 0. So this sequence is probably bounded and if it decreasing then it will converge.
If I solve for $x_n$ I get that
$x_n=4- \frac{1}{x_{n+1}}$
So now we see from this that the biggest $x_n$ could be is 4, because it seems that all our numbers are positive. I might need to use that fact that its bounded and decreasing to show that it converges.

 

[lippyka]

By the Monotone Convergence Theorem, since our sequence is bounded and decreasing, it converges. Now I just need to prove that it converges to 0. If I use the definition of this sequence and work backwards, I get: x_2=4- \frac{1}{x_3}, x_3=4- \frac{1}{x_4}, x_4=4- \frac{1}{x_5}, and so on. If I substitute all of these into the formula for x_n, I get x_n=4- \frac{1}{4-x_{n+1}} So if I solve this equation for x_{n+1} I get x_{n+1}=\frac{1}{4-x_n} which is the same formula we started with, and this shows that our sequence converges to 0. Therefore, the sequence defined by x_1=3 and x_{n+1}= \frac{1}{4-x_n} converges to 0.