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Math Help - Find the sum of the series.

  1. #1
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    Find the sum of the series.

    \sum_{k = 0}^\infty \frac{3^{k-1}}{4^{3k+1}}
    \sum_{k = 0}^\infty \frac{3^{k}*3^{-1}}{4^{3k}*4^{1}}
    \sum_{k = 0}^\infty \frac{3^{k}}{(4^{3})^{k}*4^{1}*3^{1}}
    (\frac{1}{12})\sum_{k = 0}^\infty \frac{3^{k}}{64^{k}}
    (\frac{1}{12})\sum_{k = 0}^\infty (\frac{3}{64})^{k}
    So it's a geometric series with a = 1 and r = 3/64. So \frac{1}{12}*\frac{1}{1 - \frac{3}{64}} = \frac{16}{183} ?

    The solutions manual had \frac{3}{15,616}. Am I doing this right?
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  2. #2
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    Re: Find the sum of the series.

    What you have done looks fine. Does your solutions manual post their full solution or just the answer?
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  3. #3
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    Re: Find the sum of the series.

    There was a full solution, but I only copied down \frac{3}{15,616}. I may have copied down the wrong answer... and I won't have a chance to check until tomorrow.
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  4. #4
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    Re: Find the sum of the series.

    Well, the solutions manual had:
    \sum_{k = 2}^\infty \frac{3^{k-1}}{4^{3k+1}}
    \sum_{k = 0}^\infty \frac{3^{k+1}}{4^{3k+7}}
    (\frac{3}{4^7})\sum_{k = 0}^\infty (\frac{3}{4^{3}})^k
    \frac{3}{4^7}*\frac{1}{1 - \frac{3}{4^3}}
    \frac{3}{15616}.

    But \sum_{k = 2}^\infty \frac{3^{k-1}}{4^{3k+1}} was not the original problem, \sum_{k = 0}^\infty \frac{3^{k-1}}{4^{3k+1}} was. So, pretty safe to say that's a typo?
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