Convergence of a recursive fraction

[kindlychung]

Homework Statement

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 [tex] \lim

Homework Equations

$$x_{1}=3$$ and $$x_{n+1} = \frac{1}{4-x_{n}}$$

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.

 

[mighty2000]

Thank you for your post regarding the convergence of the sequence defined by the recursive equation x_{1}=3 and x_{n+1} = \frac{1}{4-x_{n}}. I would like to provide some additional insights and explanations to your solution.

To prove the convergence of this sequence, we can use the Monotone Convergence Theorem. This theorem states that if a sequence is monotone and bounded, then it must converge. In this case, we can see that x_{n} is a decreasing sequence (since x_{n+1} is always smaller than x_{n}) and it is also bounded below by 0 (since x_{n} cannot be negative due to the denominator in the recursive equation). Therefore, by the Monotone Convergence Theorem, we can conclude that the sequence must converge.

Now, to address the second part of the problem, we need to show that if the limit of x_{n} exists, then the limit of x_{n+1} must also exist and be equal to the same value. This can be easily shown using the definition of a limit. If we assume that x_{n} converges to a limit x, then for any positive epsilon, there exists an N such that for all n > N, we have |x_{n} - x| < epsilon. Now, since n+1 > n > N, we can also say that |x_{n+1} - x| < epsilon, which means that x_{n+1} also converges to the same limit x.

Finally, in order to explicitly compute the limit of this recursive sequence, we can take the limit of both sides of the recursive equation. This will give us x = 1/(4-x), which we can solve to get the limit x = 2. Therefore, we can conclude that the limit of the sequence defined by x_{1}=3 and x_{n+1} = \frac{1}{4-x_{n}} is equal to 2.

I hope this explanation helps to clarify any confusion and provides a deeper understanding of the problem. Keep up the good work in your studies!