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Difference method with partial fractions

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Verify that, for all positive values of n,
    1/(n+2)(2n+3) -1/((n+3)(2n+5))=(4n+9)/((n+2)(n+3)(2n+3)(2n+5))
    For the series
    ∑_(n=0)^N▒(4n+9)/((n+2)(n+3)(2n+3)(2n+5))
    Find
    The sum to N terms,
    The sum to infinity.


    2. Relevant equations

    no

    3. The attempt at a solution

    i had solved the first sextion which is to verification
    however, when it came to the difference method solving, i had a huge problems
    therefore, any solutions?
     
  2. jcsd
  3. Sep 7, 2009 #2

    benorin

    User Avatar
    Homework Helper

    Google telescoping series
     
  4. Sep 7, 2009 #3
    most of the google doesnt reveal the method of solving series using difference method ==
    i have tried it
    or maybe you can give me the links if you have found it
     
  5. Sep 7, 2009 #4
    [tex]\frac{1}{(n+2)(2n+3)} = \frac{2}{2n+3} - \frac{1}{n+2}[/tex]
    [tex]\frac{1}{(n+3)(2n+5)} = \frac{2}{2n+5} - \frac{1}{n+3}[/tex]
    I think this may be the key issue bugging you when trying to use the difference method without simplifying the expressions further.
     
  6. Sep 7, 2009 #5
    well i have found the answer
    thanks for the help
    my fault is i did not classified them correctly
    as i should put 1/(n+3) and 1/(n+2) together as well as 1/(2n+3) and 2/(2n+5) together instead of i straight use the result to do difference method thats y i cant find the answer!
    Anyways thanks for your help
     
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