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Let $\{ x_n\}$ be a sequence in $\mathbb{R}$ and $x_{n+2}=\frac{1}{2}(x_{n+1}+x_{n})$. Given that $x_1=a,x_2=b$. I know the definition of convergent. But in this question, it is not obvious to apply the definition.

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up vote 1 down vote accepted

Given any $N \in \mathbb{N}$, note that for all $m, n > N$, we have $x_m$ and $x_n$ both in the interval with endpoints $X_N$ and $X_{N+1}$. The length of the intervals formed by successive elements in your sequence converges to $0$ (since it's being halved at each step) so that the sequence in question is clearly Cauchy. Recall that any Cauchy sequence of real numbers converges and you are done.

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yup, that's the idea i tried to prove but i don't know how to rewrite with the definition of cauchy sequence. – Mathematics Nov 4 '12 at 3:13
Can you be more precise about where your confusion lies? Perhaps you could edit your question to include what you have written up already. – Benjamin Dickman Nov 4 '12 at 22:02

HINT: Notice that $x_{n+2}$ is the arithmetic mean of $x_{n+1}$ and $x_n$: it is midway between $x_{n+1}$ and $x_n$. Thus, if $a<b$ we have a picture something like this (though this is not to scale):

$$\begin{array}{} a=x_0&x_2&x_4&x_6&\ldots&x_5&x_3&x_1=b \end{array}$$

If $a>b$, just turn the picture around.

Now try to prove what the diagram suggests: that $\langle x_{2n}:n\in\Bbb N\rangle$ is an increasing sequence, that $\langle x_{2n+1}:n\in\Bbb N\rangle$ is a decreasing sequence, and that $x_{2n}<x_{2m+1}$ for all $n,m\in\Bbb N$. You can also prove something useful about the sequence $\langle x_{2n+1}-x_{n2}:n\in\Bbb N\rangle$.

Now use what you know about bounded, monotone sequences in $\Bbb R$.

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Is my solution above correct? – Mathematics Nov 3 '12 at 18:55

By induction, show that $$x_n = a+ (b-a)\cdot\left(\frac23+(-1)^n\frac1{3\cdot 2^{n-2}}\right).$$ This should convince you that $x_n\to \frac{a+2b}3$.

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How is the OP supposed to justify this hypothesis before s/he tries to prove it by induction? – Fly by Night Nov 3 '12 at 22:14
(-1) Absolutely non-enlightening solution – Norbert Nov 4 '12 at 9:22
Trying a few terms, guessing the expression and proving it is an absolutely valid method. – Hagen von Eitzen Nov 4 '12 at 20:17

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