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Induction on an inequality involving sequences

  1. Nov 2, 2005 #1
    I just can't get the following question. Can someone help me out?

    Q. Let a < a_1 < b_1 and define [tex]a_{n + 1} = \sqrt {a_n b_n } ,b_{n + 1} = \frac{{a_n + b_n }}{2}[/tex] .

    a) Prove that [tex]a_n \le a_{n + 1} \le b_{n + 1} \le b_n [/tex] for all n.
    b) Deduce that the sequences {a_n} and {b_n} both converge.
    c) Prove that they have the same limit. (This limit is called the arithmetic-geometric mean of a_1 and b_1)

    Here are my attempts

    b) I'll do this one first. By the result of part 'a' {a_n} is increasing and bounded above by b_n while {b_n} is a decreasing sequence and bounded below by a_n. So by the monotonic sequence theorem both sequences converge. Not sure about the argument there, I might have missed a few important points.

    a) There might ways other than induction to do this but I can't think of any. So I need to start off with a specific case.

    For n = 1: I can get a_1 <= a_(2) but not the b_(2) <= b_(1) part 'properly' and I can't get the a_2 <= b_2 bit at all.
  2. jcsd
  3. Nov 2, 2005 #2


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    Here are a few comments. If 0< a< b then
    1) 0< a2< ab so 0< a< [tex]\sqrt{ab}[/tex]
    That gives you "a2< a1".
    2) 0< a+ b< 2b so 0< (a+b)/2< b.
    That gives you "b2< b1".
    Now, assume that [tex]a_n \le a_{n + 1} \le b_{n + 1} \le b_n [/tex]
    for some k and use the same calculations as above to show that
    [tex]a_{n+1} \le a_{n + 2} \le b_{n + 2} \le b_{n+1} [/tex]
    Last edited: Nov 3, 2005
  4. Nov 2, 2005 #3
    Thanks for the help HallsofIvy.
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