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Math Help - Evaluating this infinite sum

  1. #1
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    Evaluating this infinite sum

    Is it possible to evaluate this infinite sum:

    \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by acevipa View Post
    Is it possible to evaluate this infinite sum:

    \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)
    This way seems messy, but it works..

    Skip proving convergence.... it converges.

    \displaystyle L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)

    \displaystyle 3L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}

    \displaystyle 2L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^i

    This form is "cleaner." Write some terms

    2L = \dfrac{2}{3} + 2\left(\dfrac{4}{9}\right) + 3\left(\dfrac{8}{27}\right) + 4\left(\dfrac{16}{81}\right) + \dots

    Consider what happens when we multiply by 3,

    6L = 2 + \dfrac{2\cdot4}{3} + \dfrac{3\cdot8}{9} + \dfrac{4\cdot16}{27} + \dots

    Add the two together, matching up terms with common denominators.

    8L = 2 + \dfrac{2+2\cdot4}{3} + \dfrac{2\cdot4+3\cdot8}{9} + \dfrac{3\cdot8+4\cdot16}{27} + \dots

    = 2 + 5\left(\dfrac{2}{3}\right) + 8\left(\dfrac{2}{3}\right)^2 + 11\left(\dfrac{2}{3}\right)^3 + \dots

    So

    \displaystyle 8L = 2 + \sum_{i=1}^{\infty}(3i+2)\left(\frac{2}{3}\right)^  i

    \displaystyle = 2 + \sum_{i=1}^{\infty}3i\left(\frac{2}{3}\right)^i + 2\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^i

    Simplifying,

    8L = 2 + 3(2L) + 4

    2L = 6

    L = 3
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  3. #3
    MHF Contributor chisigma's Avatar
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    For |x|<1 is...

    \displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x} (1)

    ... so that is...

    \displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}} (2)

    Now from (2) follows...

     \displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3 (3)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    For |x|<1 is...

    \displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x} (1)

    ... so that is...

    \displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}} (2)

    Now from (2) follows...

     \displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3 (3)

    Kind regards

    \chi \sigma
    Great method. Thanks
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  5. #5
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    Quote Originally Posted by acevipa View Post
    Great method. Thanks
    If you can follow chisigma's solution you have posted this in the wrong forum and wasted the time of those helpers who have given or may have considered giving solutions that do not use calculus.

    CB
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  6. #6
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    Quote Originally Posted by acevipa View Post
    Is it possible to evaluate this infinite sum:

    \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)
    \displaystyle\ S_{\infty}=\frac{1}{3}+\left(\frac{2}{3}\right)\le  ft(\frac{2}{3}\right)+\left(\frac{3}{3}\right)\lef  t(\frac{2}{3}\right)^2+\left(\frac{4}{3}\right)\le  ft(\frac{2}{3}\right)^3+.....

    3\displaystyle\ S_{\infty}-1=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig  ht)^2+4\left(\frac{2}{3}\right)^3+....

    s=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig  ht)^2+.....

    \frac{2}{3}s=2\left(\frac{2}{3}\right)^2+3\left(\f  rac{2}{3}\right)^3+4\left(\frac{2}{3}\right)^4....

    s-\frac{2}{3}s=\frac{s}{3}=2\left(\frac{2}{3}\right)  +\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\righ  t)^3+.....

    \frac{s-2}{3}=\left(\frac{2}{3}\right)+\left(\frac{2}{3}\r  ight)^2+\left(\frac{2}{3}\right)^3+....

    which is purely a geometric series.
    Then use the result of this to calculate the original sum.
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  7. #7
    Junior Member Renji Rodrigo's Avatar
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    (just some curiosities about this type of series)
    i will use

    \Delta f(x)=f(x+1)-f(x) for the difference operator

    and \sum\limits_x f(x) = g(x) if and only if \Delta g(x) =f(x) (the indefinite summation)

    we have the summation by parts method

     \sum\limits_{x} g(x)\Delta<br />
f(x)=f(x).g(x)-\sum\limits_{x} f(x+1).\Delta g(x)

    (its similar to the integration by parts)

    if we apply this method "n" times we get
    [tex]\sum\limits_{x} g(x).\Delta f(x)=[/Math]


    \sum\limits^{n}_{k=0}\Delta^{k}g(x).(-1)^{k}\Delta^{-k}E^{k}f(x)+(-1)^{n+1}\sum\limits_{x}\Delta^{n+1}g(x)\Delta^{-n}E^{n+1}f(x)

    ( where E^k f(x)=f(x+k))

    if, g(x) has degree n , (so \Delta ^{n+1} g(x)=0 )and |a|<1 we can use it on the summation by parts to find


    \sum\limits^{\infty}_{x=c} g(x)a^{x}=a^{c-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}g(x)|_{x=c}  }{(b-1)^{k+1}}

    where b= \frac{1}{a}

    if  g(x)=x^{n} and e c=0
     \sum\limits^{\infty}_{x=0} x^{n}a^{x}=a^{-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}x^{n}|_{x=0  }}{(b-1)^{k+1}}

    but  \Delta^{k}x^{n}|_{x=0}=k!\bigg\{^{n}_{k}\bigg\} (here  \bigg\{^{n}_{k}\bigg\} it's the stirling number of the second kind, knuth's notation )
    so
     \sum\limits^{\infty}_{x=0} x^{n}a^{x}= a^{-1}\sum\limits^{n}_{k=0}\frac{k! \bigg\{^{n}_{k}\bigg\}}{(b-1)^{k+1}}.
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  8. #8
    Junior Member Renji Rodrigo's Avatar
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    (other way , the general case)

    we can define the operator xD, (xD) f(x) =x [D f(x)] ( where "D" is the derivative )
    then we can show

     (xD)^{n}g(x)=\sum\limits^{n}_{k=0}\bigg\{^{n}_{k}\  bigg\}x^{k}D^{k}g(x)

    (again the stirling numbers of the second kind)
    take the series
    \sum\limits^{\infty}_{n=0}x^{n}=\frac{1}{1-x}

    apply (xD)^p in the both sides

     \sum\limits^{\infty}_{n=0}n^{p}x^{n}=(xD)^p\frac{1  }{1-x}

    using the identity
    (xD)^{p}g(x)=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\  bigg\}x^{k}D^{k}g(x)

    we have

    (xD)^{p}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}x^{k  }D^{k}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}\fra  c{k!x^{k}}{(1-x)^{k+1}}

    so
    \sum\limits^{\infty}_{n=0}n^{p}x^{n}=\sum\limits^{  p}_{k=0}\bigg\{^{p}_{k}\bigg\}\frac{k!x^{k}}{(1-x)^{k+1}}.
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