Let $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Show that $\{a_n\}_{n=1}^{\infty}$ is Cauchy

Question

Let $a_0$ and $a_1$ be distinct real numbers. Define $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Show that $\{a_n\}_{n=1}^{\infty}$ is a Cauchy sequence. Hint: You may want to use the facts that $$a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0)$$ and $$1+x+x^2+\cdots+x^n=\frac{1-x^{n+1}}{1-x} \text{ if }x\neq 1.$$

I don't see how those two facts are helpful. I played around with them, but I didn't come up with anything useful. Thanks for your help.

Answer 1

Since you already know that $$|a_{n+1}-a_n|=\left(\frac{1}{2}\right)^n|a_1-a_0|$$,

$$|a_m-a_n|\le|a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+....+|a_{n+1}-a_n|$$, where $m \gt n$. Hence using the first equality we get $$|a_m-a_n|\le |a_1-a_0|\left[\left(\frac{1}{2}\right)^{m-1}+\left(\frac{1}{2}\right)^{m-2}+....+\left(\frac{1}{2}\right)^{n}\right]$$

$$ \le |a_1-a_0|\left(\frac{1}{2}\right)^n\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+....\right]=|a_1-a_0|\left(\frac{1}{2}\right)^{n-1}$$

Now you can do it

Answer 2

Note that $$|a_n-a_m|\leqslant |a_n-a_{n+1}+a_{n+1}-\cdots+a_{m-1}-a_m|$$ then use the triangle inequality.

Answer 3

Here is my trial. Because \begin{align*} a_{n+2}-a_{n+1}&=\frac{a_{n+1}+a_n}{2}-a_{n+1}\\ &=\frac{a_n-a_{n+1}}{2}, \forall n\geq 1, \end{align*} we get $$|a_{n+2}-a_{n+1}|=\frac{1}{2}|a_{n+1}-a_{n}|, \forall n\geq1.$$ Now applying the result of Line 4 of Page 74 in Apostol's book, Mathematical Analysis, second edition, we see that $a_n$ is convergent.

Now let's try to estimate that limit. Since $a_{n+1}=\frac{a_n+a_{n-1}}{2}, $ add $a_n/2$ to both sides of this equality, we get \begin{align*} a_{n+1}+\frac{a_n}{2}&=\frac{a_n+a_{n-1}}{2}+\frac{a_n}{2}\\ &=a_{n}+\frac{a_{n-1}}{2}, \end{align*} which implies, after some simple iterations, that \begin{align*} a_{n+1}+\frac{a_n}{2}&=a_n+\frac{ a_{n-1}}{2}\\ &=a_{n-1}+\frac{a_{n-2}}{2}\\ &=\cdots\\ &=a_1+\frac{a_0}{2}. \tag{$*$} \end{align*} Let $A=\lim_{n\to\infty} a_n,$ we get by equation $(*)$ that $$\lim_{n\to\infty} a_n=\frac{2a_1}{3}+\frac{ a_0}{3}.$$