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Comparison Test and Geometric Series

  1. Nov 23, 2012 #1
    The Problem:

    Let a(n) = (n^2)/(2^n)

    Prove that if n>=3, then:

    (a(n+1))/(a(n)) <= 8/9

    By using this inequality for n = 3,4,5,..., prove that:

    a(n+3) <= ((8/9)^n)(a(3))

    Using the comparison test and results concerning the convergence of the geometric series, show that:

    The SUM(from n=1 to infinity) of a(n+3) is convergent.

    Now use the shift rule to show that:

    The SUM(from n=1 to infinity) of a(n) is convergent.

    Attempt:

    Shown that if you let n=3 then (a(n+1))/(a(n)) = 8/9

    How do I then show that for all n>=3, (a(n+1))/(a(n)) <= 8/9

    From then on I'm not quite sure :(

    Thanks for any help and sorry for the lack of LaTex on a phone haha
     
    Last edited: Nov 23, 2012
  2. jcsd
  3. Nov 23, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Your "attempt" seems to be missing something- like any attempt at all! There are a number of ways to show that "a< b". One is to show that b/a> 1. another is to show that b-a> 0.
     
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