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Math Help - Geometric series

  1. #1
    Junior Member
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    Geometric series

    Not sure about this one:

    Find the sum of the infinite geometric series 1+3/5+9/25+27/125+...

    Thanks for any help.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mike1 View Post
    Not sure about this one:

    Find the sum of the infinite geometric series 1+3/5+9/25+27/125+...

    Thanks for any help.
    In geometric sequence there is a common ratio, r, between two successive terms.
    a2 = a1 *r
    a3 = a2 *r = a1 *r^2
    ....
    an = a1 *r^(n-1)

    In a geometric series up to infinity, if the absolute value of r is less than 1, then the series converges and its sum is a1/(1-r).

    1 +3/5 +9/25 +27/125 +...

    r = (3/5)/1 = 3/5
    r = (9/25)/(3/5) = (9/25)*(5/3) = 3/5
    r = (27/125)/(9/25) = (27/125)*(25/9) = 3/5
    Ok, r = 3/5 = 0.6 ----less than 1, so there is a sum.

    Sum = a1 / (1-r)
    Sum = 1 / (1 -3/5) = 1/(2/5) = 5/2 -------------answer.
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  3. #3
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    Hello, Mike!

    There is a formula for the sum of an infinite geometric series.

    . . . . . . . . a
    . . S . = . ------
    . . . . . . . 1 - r

    where a = first term, r = common ratio.


    Find the sum of the infinite geometric series: 1 + 3/5 + 9/25 + 27/125 + ...

    We have: a = 1, r = 3/5

    . . . . . . . . . . . .1 . . . . . . 1 . . . . 5
    Then: . S .= .--------- .= .----- .= . --
    . . . . . . . . . .1 - 3/5 . . . 2/5 . - - -2

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