# Cauchy sequence question

by Unassuming
Tags: cauchy, sequence
 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?
 HW Helper Sci Advisor Thanks P: 24,421 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.
HW Helper
 PF Patron Sci Advisor Thanks Emeritus P: 38,387 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!