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Math Help - sum of series

  1. #1
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    sum of series

    i hope this is the right place:

    Find the sum of:
     \sum_{n=1} ^{10} {3*(\frac{1}{4})^n}

    =1-(\frac{1}{4})^{10}) = 1

    but there is an additional problem:

    Find the error between S_{10} and s_\infty. Use the formula for  S_n and  s_\infty of convergement geometric series , find a general formula that gives the error between  S_n and  s_\infty. Use the new formula to explain why, for the series in #7, the error between  S_n and  s_\infty is always equal to r^n.

    I don't understand the problem. please help me out! thanks!
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  2. #2
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    Quote Originally Posted by Anemori View Post
    i hope this is the right place:

    Find the sum of:
     \sum_{n=1} ^{10} {3*(\frac{1}{4})^n}

    =1-(\frac{1}{4})^{10}) = 1

    but there is an additional problem:

    Find the error between S_{10} and s_\infty. Use the formula for  S_n and  s_\infty of convergement geometric series , find a general formula that gives the error between  S_n and  s_\infty. Use the new formula to explain why, for the series in #7, the error between  S_n and  s_\infty is always equal to r^n.

    I don't understand the problem. please help me out! thanks!
    \sum_{k=0}^nar^k=a\frac{1-r^{n+1}}{1-r}

    \sum_{k=0}^\infty ar^k=\frac{a}{1-r}

    Since we are dealing with sums starting with k=1, we'll need to subtract that out.

    S_n=\sum_{k=1}^nar^k=a\frac{1-r^{n+1}}{1-r}-a

    s_\infty = \sum_{k=1}^\infty ar^k=\frac{a}{1-r}-a

    The term error is used because the sequence S_1, S_2, S_3, \cdots can be seen as approximations to s_\infty that get closer and closer as n increases. The error is the true value minus the value of the approximation, or s_\infty-S_n.
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