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Cauchy sequence question 
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#1
Oct1608, 04:09 PM

P: 169

Let a_1 and a_2 be arbitrary real number that are not equal. For [tex]n \geq 3[/tex], define a_n inductively by,
[tex]a_n = \frac{1}{2} (a_{n1}+a_{n+2} )[/tex] I cannot get the result that the book gets. I proceed, [tex]a_{n+1}  a_{n} = \frac{1}{2}(a_n + a_{n1} )  \frac{1}{2} (a_{n1} + a_{n2} ) = \frac{1}{2} ( a_n  a_{n2} )= \frac{1}{2}(a_n + a_{n1} ) [/tex] The book got the answer, [tex] a_{n+1}  a_n = \frac{1}{2} (a_n  a_{n1} ) [/tex] Any help for me? 


#2
Oct1608, 04:28 PM

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You've got some typos in there that are making this really confusing. The original recursion is a_n=(1/2)*(a_n1+a_n2), right? That says 2*a_na_n1=a_n2. I agree with the book.



#3
Oct1608, 05:19 PM

P: 169

I can't see it. I've done some algebra and I still can't get anywhere from your last step.



#4
Oct1608, 05:24 PM

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Cauchy sequence question
You've got a_n+1a_n=(1/2)(a_na_n2), right? Put the expression for a_n2 into that. If you still aren't getting it show your work.



#5
Oct1608, 05:40 PM

P: 169

I got it. Thanks Dick



#6
Oct1708, 07:12 AM

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Here's a "cheat":
Since [itex]\sum x^n/n![/itex] is the Taylor's series for e^{x}, 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! 


#7
Oct1708, 03:35 PM

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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