# Convergence of a recursive fraction

1. Sep 26, 2010

### kindlychung

1. The problem statement, all variables and given/known data
1. Prove that the sequenced defined by $$x_{1}=3$$ and $$x_{n+1} = \frac{1}{4-x_{n}}$$ converges.
2. Now that we know $$\lim x_{n}$$ exists, explain why $$\lim x_{n+1}$$ must exist and equal the same value.
3. Take the limit of each side of the recursive equation in part 1 of this exercise to explicitly compute $$\lim 2. Relevant equations [tex]x_{1}=3$$ and $$x_{n+1} = \frac{1}{4-x_{n}}$$

3. The attempt at a solution
1. If we try to crunch each term of $$x_{n}$$ starting from $$x_{1}$$, we can easily see that $$x_{n}$$ is monotone and bounded, hence convergent. The the recursive form of fraction is rather annoying, makes it hard to prove.

2. Let $$\lim x_{n} = x$$, then there is an N, such that whenever $$n \ge N$$, we have $$|x_{n} - x | < \epsilon$$ for any positive $$\epsilon$$, but n+1 > n > N, hence $$|x_{n+1} - x | < \epsilon$$, as desired.

3. Let $$\lim x_{n} = \lim x_{n+1} = x$$, take limits of both sides, we have $$x = \frac{1}{4-x}$$, solve the equation we get the limit x.