Results 1 to 10 of 10

Math Help - series, please help

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    29

    series, please help

    a convergent series with x<1, and a, b are constants:

    \sum_{i=a}^{inf}  \frac{x^i}{1+ i \cdot b}

    what is and how to find its closed form ???

    please help, many thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by graticcio View Post
    a convergent series with x<1, and a, b are constants:

    \sum_{i=a}^{inf}  \frac{x^i}{1+ i \cdot b}

    what is and how to find its closed form ???

    please help, many thanks!
    I am going to assume a=0. Say that b=1.

    We want to sum, \sum_{k=0}^{\infty} \frac{x^k}{k+1}

    Consider \sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}.

    That means, \int_0^t \sum_{k=0}^{\infty} x^{k} dx = \int_0^t \frac{1}{1-x} dx

    Thus, \sum_{k=0}^{\infty} \int_0^t x^{k} dt = - \ln (1-t) for 0<t<1

    Thus, \sum_{k=0}^{\infty} \frac{t^{k+1}}{k+1} = - \ln (1-t)

    Thus, \sum_{k=0}^{\infty} \frac{t^k}{k+1} = - t^{-1} \ln (1-t)

    ----
    In the general case consider, \sum_{k=0}^{\infty} x^{bk} = \frac{1}{1-x^b}
    And apply a similar argument.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2008
    Posts
    29
    thanks for your reply

    i hav tried a similar approach
    but when a \neq 0

    we hav to start with
     \frac{x^{ab}}{1 - x^b}

    and the integration of it
     \int \frac{x^{ab}}{1 - x^b} dx
    i hav checked with some integration tables
    but there is only reduction formula available

    i wonder how to get a "more closed" form of the integral
    or if there is any other way to find the sum of tht series
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by graticcio View Post
    a convergent series with x<1, and a, b are constants:

    \sum_{i=0}^{\infty}  \frac{x^i}{1+ i \cdot b}

    what is and how to find its closed form ???
    \sum\limits_{i = 0}^\infty  {\frac{{x^i }}<br />
{{1 + bi}}}  = \sum\limits_{i = 0}^\infty  {x^i \int_0^1 {u^{bi} \,du} }  = \int_0^1 {\left[ {\sum\limits_{i = 0}^\infty  {\left( {xu^b } \right)^i } } \right]\,du}  = \int_0^1 {\frac{1}<br />
{{1 - xu^b }}\,du} .

    There's your closed form.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2008
    Posts
    29
    thanks

    what if the series does not start with 0
    but some constant a ???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by graticcio View Post
    a convergent series with x<1, and a, b are constants:

    \sum_{i=a}^{inf}  \frac{x^i}{1+ i \cdot b}

    what is and how to find its closed form ???

    please help, many thanks!
    As an entree:

    For the special case a = 0 and b = 1, it's not hard to use the standard result for an infinite geometric series to show that the series is equal to -\frac{\ln (1 - x)}{x}.
    Last edited by mr fantastic; January 19th 2008 at 07:01 PM. Reason: Haven't edited, but just saw all the preceeding posts. The things that happen when you start something and then wander off ..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    As an entree:

    For the special case a = 0 and b = 1, it's not hard to use the standard result for an infinite geometric series to show that the series is equal to -\frac{\ln (1 - x)}{x}.
    But I see Krizalid's skipped the soup du jour and gone to the main course. That just leaves desert:

    Quote Originally Posted by graticcio View Post
    thanks

    what if the series does not start with 0
    but some constant a ???
    Then you just relabel the dummy index ......

    \sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b} = \sum_{n=0}^{inf} \frac{x^{n+a}}{1+ (n+a) \cdot b} = .....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2008
    Posts
    29
    thanks

    yes, shifting the index, but then it comes back
    to the problem i mentioned in the second reply

    \sum\limits_{n = 0}^\infty {\frac{{x^{n+a}}} {{1 + b(n+a)}}}= \int_0^1 {\left[ {\sum\limits_{n = 0}^\infty {\left( {xu^b } \right)^{n+a} } } \right]\,du} = \int_0^1 {\frac{(xu^b)^a} {{1 - xu^b }}\,du}

    cannot find the 'closed' form of that integral
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by graticcio View Post
    thanks

    yes, shifting the index, but then it comes back
    to the problem i mentioned in the second reply

    \sum\limits_{n = 0}^\infty {\frac{{x^{n+a}}} {{1 + b(n+a)}}}= \int_0^1 {\left[ {\sum\limits_{n = 0}^\infty {\left( {xu^b } \right)^{n+a} } } \right]\,du} = \int_0^1 {\frac{(xu^b)^a} {{1 - xu^b }}\,du}

    cannot find the 'closed' form of that integral
    If you enter the integrand into the Wolfram integrator (make sure you use x, not u as your integration variable), you'll get a closed form for the indefinite integral. Now evaluate the definite integral.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jan 2008
    Posts
    29
    thanks
    Last edited by graticcio; January 20th 2008 at 05:58 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 02:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 07:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  4. Replies: 2
    Last Post: December 1st 2009, 01:45 PM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum