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Math Help - sequences- covergent?

  1. #1
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    sequences- covergent?

    {a_{n + 1}} = \sqrt {2{a_n}} <br />
    Is this a convergent sequence and if so, what does it converge to?


    I have no idea how to beginning this...
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    {a_{n + 1}} = \sqrt {2{a_n}} <br />
    Is this a convergent sequence and if so, what does it converge to?


    I have no idea how to beginning this...
    What is the first term?



    Graphing some terms for a_n = 1 it looks like it converges to 2.
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  3. #3
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    I wasn't given a first term which is why I am having I problem

    but {a_n} = 1?
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  4. #4
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    It is easy to see that the series converges for any value of a_1>0
    if a_1=0 then the series converges to zero.

    Now lets assume that a_n>2 then it follows that:

    a_{n+1}=\sqrt{2a_n}<\sqrt{a_n\cdot a_n}=a_n

    And

    a_{n+1}=\sqrt{2a_n}>\sqrt{2\cdot2}=2

    So you see that if a_n>2 then: 2<a_{n+1}<a_n.

    Now what happens if a_n<2 ??

    Hope that helps.
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