# Cauchy sequence question

by Unassuming
Tags: cauchy, sequence
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 P: 169 Let a_1 and a_2 be arbitrary real number that are not equal. For $$n \geq 3$$, define a_n inductively by, $$a_n = \frac{1}{2} (a_{n-1}+a_{n+2} )$$ I cannot get the result that the book gets. I proceed, $$a_{n+1} - a_{n} = \frac{1}{2}(a_n + a_{n-1} ) - \frac{1}{2} (a_{n-1} + a_{n-2} ) = \frac{1}{2} ( a_n - a_{n-2} )= \frac{1}{2}(a_n + a_{n-1} )$$ The book got the answer, $$a_{n+1} - a_n = \frac{-1}{2} (a_n - a_{n-1} )$$ Any help for me?
 Sci Advisor HW Helper Thanks P: 25,235 You've got some typos in there that are making this really confusing. The original recursion is a_n=(1/2)*(a_n-1+a_n-2), right? That says 2*a_n-a_n-1=a_n-2. I agree with the book.
 P: 169 I can't see it. I've done some algebra and I still can't get anywhere from your last step.
 Sci Advisor HW Helper Thanks P: 25,235 Cauchy sequence question You've got a_n+1-a_n=(1/2)(a_n-a_n-2), right? Put the expression for a_n-2 into that. If you still aren't getting it show your work.
 P: 169 I got it. Thanks Dick
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Here's a "cheat": Since $\sum x^n/n!$ is the Taylor's series for ex, 1+ 1/2!+ 1/3!+ ... converges to e. Since it converges, it is a Cauchy sequence! Or you could just note that the numeric series converges by the ratio test. Again, since it converges, it is a Cauchy sequence. But I doubt that either of those answers is what was wanted!
 P: 169 Those are nice things to put the problem into perspective though. Sometimes I (and hopefully others) get lost in the algebra of the problem.

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