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Q about sequences

  1. Mar 4, 2008 #1
    The sequence {an} is defined by

    a1=1,an+1=[tex]\sqrt{1+an/2}[/tex] ,n=1,2,3,4,...

    (a) a^2n-2<0 , (b)a^2a+1-a^2n>0. deduce that{an} converges and find its limits?

    please help me get the answer................
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2


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    Are the (a) and (b) [itex]a^{2n-2}[/itex] and [itex]a^{2n-1}- a^{2n}> 0[/itex]? If so, what is "a" and what do they have to do with the given sequence?

    Have you never seen any sequences like this before?? You don't seem to have even tried anything. While looking at a few terms won't prove anything, it might help you see what's happening: [itex]a_1= 1[/itex], [itex]a_2= \sqrt{1+ 1/2}= \sqrt{3/2}[/itex], [itex]a_3= \sqrt{1+ \sqrt{3/2}/2}= \sqrt{(2+ \sqrt{3/2})/2}[/itex]. Does that look like it is increasing? Do you know any theorem about convergence of increasing sequences?

    As for finding the limit, what happens if you take the limit on both sides of the equation [itex]a_{n+1}= \sqrt{1+ a_n/2}[/itex]?
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