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Sum of convergent series HELP

  1. Apr 27, 2009 #1
    Sum of convergent series HELP!!

    1. The problem statement, all variables and given/known data
    Find the sum of the convergent series -
    the sum of 6 / (n+7)(n+9) from n=1 to infinity (∞)

    A) 31/24
    B) 45/56
    C) 8/11
    D) 17/24
    E) 23/24


    2. The attempt at a solution
    I was looking in the book and they had one example that was kinda close so i tried to follow it and i did this....

    = 6 times the sum of (1/(n+7))-(1/n+9))
    = 6 times (1/8-1/10)+(1/9-1/11)... and so on

    but that got me nowhere.

    I tried plugging this into my calculator sum(seq(6/(x+7)(x+9),x,1,999,1)
    and i got .7023779 which is the closest to answer D --> .70833333
    but that's only up to term 999 and how do i know that the series won't get bigger than 17/24?

    i found that up to the 7,490th term the sum is .7075417. which is closer so i'm going to go with answer D. :)
     
    Last edited: Apr 27, 2009
  2. jcsd
  3. Apr 27, 2009 #2

    Mark44

    Staff: Mentor

    Re: Sum of convergent series HELP!!

    Of the million ways you've tried, show us your work for one of them, and we'll give it our best.
     
  4. Apr 27, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Sum of convergent series HELP!!

    My thought would be to start with "partial fractions".
    6/(x+9)(x+7)= A/(x+9)+ B/(x+7) so, multiplying by (x+9)(x+7), 6= A(x+7)+ B(x+9). If x= -7, 6= A(0)+ B(2) so B= 3. If x= -9, 6=A(-2)+ B(0) so A= -3:
    6/(x+9)(x+7)= 3/(x+7)- 3/(x+9)= 3(1/(x+7)- 1/(x+9))
    Sum 6/(x+9)(x+7)= 3[ sum(1/(x+7))- sum(1/(x+9)].
    This gives a "telescoping series".
     
  5. Apr 27, 2009 #4
    Re: Sum of convergent series HELP!!


    Telescoping series -
    (b sub 1 - b sub 2)+(b sub 2 - b sub 3)+(b sub 3 - b sub 4)+ ...
    Sum = (b sub 1) - lim (b sub n+1)

    So with the telescoping series I get

    3[ lim (1/8) - 1/(x+9)]

    But that's not right.... umm....
    are you sure that is a telescoping series?
    (1/8 - 1/10)+(1/9 - 1/11)+(1/10 - 1/12)...
    i thought the last # in the parenthesis has to be the same as the first # in the next parenthesis.
     
  6. Apr 28, 2009 #5

    Mark44

    Staff: Mentor

    Re: Sum of convergent series HELP!!

    No, that doesn't have to be the case.
    Your series in expanded form looks like this:
    3[(1/8 - 1/10)+(1/9 - 1/11)+(1/10 - 1/12) + (1/11 - 1/13) + (1/12 - 1/14) + ... + (1/(n + 7) - 1/(n + 9) + ...]

    You should notice that the only terms that are left after doing the subtractions are 1/8, 1/9, and -1/(n + 9), so the sequence of partial sums looks like this:
    Sn = 3(1/8 + 1/9 - 1/(n + 9))
    As n grows large without bound, what does Sn approach?
     
  7. Apr 28, 2009 #6
    Re: Sum of convergent series HELP!!

    oh so it would be 3(17/72)= 17/24
    i get it.. Thank you very much!
     
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