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Math Help - sum of infinite series

  1. #1
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    sum of infinite series

    this is for my stats class and i dont have a calc book with me...

    the sum of 1/(3^x) where x is from 1 to infinity

    thanks
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  2. #2
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    \frac{a}<br />
{{1 - r}} = \frac{{\frac{1}<br />
{3}}}<br />
{{1 - \frac{1}<br />
{3}}}
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  3. #3
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    what do a and r represent, and why are they both 1/3 in this case? thanks
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  4. #4
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    a is the first term of the series, and r is the common ratio (the ratio between consecutive terms).
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  5. #5
    Super Member 11rdc11's Avatar
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    This is a geometric series so you use the formula plato gave to find the sum
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  6. #6
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    ok what if the ratio is 5x/6, is there another formula i can use?

    this is for solving the sum of:

    x * (1/6) * (5/6)^(x-1)

    where x is 1 to infinity
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  7. #7
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    \sum\limits_{k = J}^\infty  {ax^k }  = \frac{{ax^J }}<br />
{{1 - x}},\,\,\left| x \right| < 1
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  8. #8
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    Quote Originally Posted by Plato View Post
    \sum\limits_{k = J}^\infty  {ax^k }  = \frac{{ax^J }}<br />
{{1 - x}},\,\,\left| x \right| < 1

    does a = x*(1/6) then?

    that wont work because then the answer still has an x in it.
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  9. #9
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    Quote Originally Posted by Dubulus View Post
    ok what if the ratio is 5x/6, is there another formula i can use? this is for solving the sum of:
    x * (1/6) * (5/6)^(x-1) where x is 1 to infinity
    To be honest, I cannot say that I really follow what you are trying to do.
    Here is an example: \sum\limits_{x = 1}^\infty  {\left( {\frac{5}{6}} \right)^x }  = \frac{{\frac{5}{6}}}{{1 - \frac{5}{6}}}
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  10. #10
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    Quote Originally Posted by Dubulus View Post
    ok what if the ratio is 5x/6, is there another formula i can use?

    this is for solving the sum of:

    x * (1/6) * (5/6)^(x-1)

    where x is 1 to infinity
    The correct formula for this particular sum is \frac{a}{(1-r)^2}.
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  11. #11
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    Quote Originally Posted by icemanfan View Post
    The correct formula for this particular sum is \frac{a}{(1-r)^2}.

    my problem is that there is an "x" in the sum. this to me means that each new factor in the series isn't changing by a constant ratio.
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