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Math Help - limit of a sequence

  1. #1
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    limit of a sequence

    Hey everyone.
    I wanted to get help finding the limit of the following sequence (n->inf):
    lim(1/(2^n + 1) + 2/(2^n + 2) + 4/(2^n + 3) + .. + 2^(n-1)/(2^n + n))

    I am having trouble because the number of operands to add depends on n.

    Thanks a lot,
    Michael.
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  2. #2
    Junior Member Kanwar245's Avatar
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    Re: limit of a sequence

    Is this the question you have?
    \lim_{n\rightarrow\infty} \frac{1}{2^{n+1}} + \frac{2}{2^{n+2}} + \frac{4}{2^{n+3}} + ... + \frac{2^{n-1}}{2^{n+n}}
    Last edited by Kanwar245; March 16th 2013 at 10:58 AM.
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  3. #3
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    Re: limit of a sequence

    Exactly, Thanks
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  4. #4
    Junior Member Kanwar245's Avatar
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    Re: limit of a sequence

    is the last term right, from the pattern I see it seems there's a typo for the last term?

    EDIT: Nevermind!
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  5. #5
    Junior Member Kanwar245's Avatar
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    Re: limit of a sequence

    Okay so the nth term is a_{n} = \frac{2^{n-1}}{2^{2n}} which simplifies to
    a_{n} = \frac{1}{2^{n-1}}

    so you have \lim_{n\rightarrow\infty}a_{n}

    Do you think you can do it now?
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  6. #6
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    Re: limit of a sequence

    I just took another look at the question and it seem I made a mistake.
    The +n in the denominator is not for the power, it is just +n.
    So the nth term would be 2^(n-1)/((2^n) + n)
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