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This is a practice problem for a midterm in a real analysis undergrad class, but I tagged it as homework anyway.

Let $x_0\in R$ and define $x_{n+1}=\frac{1}{2}(3-x_n)$ for $n\geq 0$. Prove that $x_n\to 1$ as $n\to \infty$.

To start I know that if $x_n$ is convergent then for sufficiently large $k$, $x_k$ is within the epsilon neighborhood of $x_{k+1}$. So if $x_n$ is convergent the limit is the solution to $C=\frac12(3-C)$ or $C=1$. Now if I can prove $x_n$ is convergent, it must converge to $1$, but I cannot figure out how to prove it. Any help would be great. My midterm is tomorrow unfortunately.

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David, I changed your $X$’s to $x$’s to be consistent with the problem statement. –  Brian M. Scott Nov 17 '11 at 9:55
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3 Answers

up vote 4 down vote accepted

Here’s a simpler version of Paul’s approach. You know that you want the limit to be $1$, so look at the discrepancies $|x_n-1|$ between $x_n$ and $1$:

$$\left|\frac{x_{n+1}-1}{x_n-1}\right|=\left|\frac{\frac12(3-x_n)-1}{x_n-1}\right|=\left|\frac{1-x_n}{2(x_n-1)}\right|=\frac12\;,$$ so $$\lim_{n\to\infty}(x_n-1)=0\;.$$

Added: By the way, the first sentence after the problem statement is true, because a convergent sequence is Cauchy, but it isn’t what you want in order to justify what follows. You want to say that if the $x_n$ converge to some $C$, then for sufficiently large $k$, $x_k$ is within epsilon of $C$ and argue from there. Alternatively, you can simply argue that if $$\lim_{n\to\infty}x_n=C\;,$$ then $$C=\lim_{n\to\infty}x_n=\lim_{n\to\infty}\frac12(3-x_n)=\frac32-\frac12\lim_{n\to\infty}x_n=\frac32-\frac{C}2$$ and continue as before.

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By definition, we have $$|x_{n+2}-x_{n+1}|=\left|\frac{1}{2}(3-x_{n+1})-\frac{1}{2}(3-x_{n})\right|=\frac{1}{2}|x_{n+1}-x_{n}|.$$ From this, the sequence $\{x_n\}$ is a Cauchy sequence: $$|x_{m}-x_{n}|=\left|\sum_{k=n}^{m-1}(x_{k+1}-x_{k})\right|\leq\sum_{k=n}^{m-1}|x_{k+1}-x_{k}|= \sum_{k=n}^{m-1}\frac{1}{2^{k}}|x_{1}-x_{0}|\rightarrow 0$$ as $n\rightarrow\infty$. Therefore, it's convergent.

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+1 for Cauchy sequence, while maybe I myself will apply a method similar to Brian's. –  puresky Nov 17 '11 at 10:04
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Consider the map $$f(x)=\frac1 2\cdot(3-x_n)$$ We can recover your sequence by setting $$x_{n+1}=f(x_n)$$ It can easily be seen that $|f(x)-f(y)|=\frac1 2|x-y|$, which means that it is a contraction. Therefore you know by the Banach fixed point theorem that this function has exactly one fixed point and your sequence converges towards that fixed point. You have already calculated that this unique fixed point is $1$.

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It’s fairly likely that at this point in the course David doesn’t have the Banach fixed point theorem as available as a tool. Fortunately, since he knows what the limit should be, he can still use the contractive nature of the map to get the desired result. –  Brian M. Scott Nov 17 '11 at 9:48
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