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Math Help - Miscellaneous series

  1. #1
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    Unhappy Miscellaneous series

    Find the sum of the infinite series
    1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3............ .............
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matsci0000 View Post
    Find the sum of the infinite series
    1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3............ .............
    Note that this is the same as \sum_{k=0}^{\infty}\left[\sum_{j=0}^{k}a^{j}\cdot r^k\right].

    If \left|a\right|<1 and \left|r\right|<1, then it follows that

    \sum_{k=0}^{\infty}\left[\sum_{j=0}^{k}a^{j}\cdot r^k\right]=\sum_{k=0}^{\infty}\left(\frac{1-a^{k+1}}{1-a}\right)r^k=\frac{1}{1-a}\sum_{k=0}^{\infty}r^k-a\left(ar\right)^k =\frac{1}{1-a}\left[\frac{1}{1-r}-\frac{a}{1-ar}\right]=\frac{1}{1-a}\left[\frac{1-ar-a+ar}{\left(1-r\right)\left(1-ar\right)}\right]=\frac{1}{\left(1-r\right)\left(1-ar\right)}
    Last edited by Chris L T521; August 23rd 2009 at 10:28 AM. Reason: fixed major mistake
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post

    If \left|a\right|<1 and \left|r\right|<1, then it follows that
    How can you make this assertion though? Is this assumed somehow? The OP didn't include any restrictions on a or r.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by QM deFuturo View Post
    How can you make this assertion though? Is this assumed somehow? The OP didn't include any restrictions on a or r.
    I imposed those conditions in order to evaluate the [geometric] series. Otherwise you can't [It diverges]!
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  5. #5
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    Hello, matsci0000!

    Find the sum: . S \;=\;1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3 + \hdots
    Multiply by (1-a)\!:

    . . (1-a)S \;=\;(1-a) + (1-a^2)r + (1-a^3)r^2 + (1-a^4)r^3 + \hdots

    . . (1-a)S \;=\;(1 - a) + (r - a^2r) + (r^2 - a^3r^2) + (r^3 - a^4r^3) + \hdots

    . . (1-a)S \;=\;\underbrace{(1 + r + r^2 + r^3 + \hdots)} - a\underbrace{(1 + ar + a^2r^2 + a^3r^3+ \hdots)}
    . . . . . . . . . . . . . . . . . . . ^{\nwarrow\text{ geometric series }\nearrow}

    . . (1-a)S \;=\;\frac{1}{1-r} - a\left(\frac{1}{1-ar}\right)

    . . (1-a)S \;=\;\frac{1-a}{(1-r)(1-ar)}

    Therefore: . S \;=\;\frac{1}{(1-r)(1-ar)}

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