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A recursive root convergions question

  1. Dec 11, 2008 #1
    i wrote the question
    and how i tried to solve it this link

    http://img514.imageshack.us/img514/5053/img9151yd0.gif [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 11, 2008 #2


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    You are given that [itex]a_1= 1[/itex] and [itex]a_{n+1}= \sqrt{x+ a_ny}[/itex] for fixed positive numbers x and y. You are trying to find prove that the sequence converges and find its limit, right?

    You then write that you are using
    [tex]L= \lim_{n\rightarrow\infty} ^n\sqrt{a_n}= \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}[/tex]
    I have no idea why or why you think those two limits are the same! I suspect you have combined the ratio test and the root test. It is true that if either of those limits is less than 1 then the sequence converges but they are not necessarily equal.

    In any case, that will not tell you what the limit of the sequence is and there is a much easier way to find it. First, although you will need to prove the sequence converges, start by assuming the sequence converges to "a".

    Now take the limit on both sides of the defining equation: look at [itex]\lim_{n\rightarrow\infty} a_{n+1}= [itex]lim_{n\rightarrow\infty}\sqrt{x+ a_ny}[/itex].
    That should easily give you an equation for a in terms of x and y.

    Is that limit less than or greater than 1 (that may depend on x and y)? If less than 1, try to prove this is a decreasing sequence having a lower bound. If greater than 1, try to prove it is an increasing sequence having an upper bound.
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