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let $a_1=1$, $a_2=2$, and define $a_n=\frac{1}{2}(a_{n-1}+a_{n-2})$. How can I show that this sequence is Cauchy?

I began with $|a_n-a_{n-1}|=\frac{1}{2}(a_n-a_{n-1})$

which goes to $\frac{1}{2}(a_{n-1}+a_{n-2})-\frac{1}{2}(a_{n-2}+a_{n-3})$

I am have been toying with this for quite some time and it is just not clicking. Can somebody please help!

Kyle H.
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3 Answers3

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Try to prove the following: $|a_n-a_{n-1}|=1/2^{n-1}$.

If you prove that lemma, then for fixed $\epsilon>0$, choose an $N$ so that $1/2^{N-1}<\epsilon$.

Then, if $m,n\geq N$ (wlog say $m\geq n$), you have $$|a_m-a_n|\leq |a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+...+|a_{n+1}-a_n|$$ $$=\frac{1}{2^{m-1}}+...+\frac{1}{2^n}=\frac{1}{2^n}(\frac{1}{2^{m-1-n}}+...+1)<\frac{1}{2^n}(2)=\frac{1}{2^{n-1}}$$but since $n\geq N$, we have that $\frac{1}{2^{n-1}}\leq \frac{1}{2^{N-1}}<\epsilon$.

Daniel Montealegre
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  • thanks for this! Seeing the triangle inequality pop out helped a lot – Kyle H. Oct 04 '13 at 02:09
  • Quick question however. Without this lemma, would I have to just look at the terms of the sequence and determine that $|a_n-a_{n-1}|=\frac{1}{2^{n-1}$? – Kyle H. Oct 04 '13 at 02:50
  • @KyleH. Are asking how did I come up with that? If so, I just saw the difference of the first couple of terms and realized that there were negative powers of $2$> – Daniel Montealegre Oct 04 '13 at 03:09
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Do you notice that the distances between succeeding terms decay exponentially?

ncmathsadist
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You said yourself that

$|a_n-a_{n-1}| = \frac{1}{2}(a_{n-1}+a_{n-2}) - \frac{1}{2}(a_{n-2}+a_{n-3})$.

Cancelling the $a_{n-2}$ terms leaves $\frac{1}{2}(a_{n-1}-a_{n-3})$. Expanding $a_{n-1}$ gives

$\frac{1}{2}(a_{n-1}-a_{n-3}) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}+a_{n-3})-a_{n-3})\right) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}-a_{n-3}))\right)$.

This means $(a_n-a_{n-1})$ = $\frac{1}{4} (a_{n-2} - a_{n-3})$. If $a_1$ and $a_2$ are fixed then $a_n-a_{n-1}$ follows exponential decay and converges to $0$.

Leo Izen
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