1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please help me solve this simple infinite sum

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    s=[itex]\sum^{k=\infty}_{k=1}[/itex] [itex]\frac{2}{k^{2}+10k+24}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I can pull the 2 out. I can set n=[itex]\infty[/itex] and just take the limit as n approaches infinity of the sum from 1 to n.

    I use partial fractions to write out:

    [itex]\frac{1}{(n+4)(n+6)}[/itex]=[itex]\frac{1}{2n+8}[/itex]-[itex]\frac{1}{2n+12}[/itex]

    To see it better, I write out the first few terms manually. I get the following (as n approaches infinity of course):

    [itex]\frac{s}{1}[/itex]=[itex]\frac{1}{5}[/itex]-[itex]\frac{1}{7}[/itex]+[itex]\frac{1}{6}[/itex]-[itex]\frac{1}{8}[/itex]+[itex]\frac{1}{7}[/itex]-[itex]\frac{1}{9}[/itex]+[itex]\frac{1}{8}[/itex]-[itex]\frac{1}{10}[/itex]+...+[itex]\frac{2}{2n+8}[/itex]-[itex]\frac{2}{2n+12}[/itex]

    But I am very confused. It is similar to a telescoping series.. but I am not used to something like this. Usually the 2nd term cancels out the third, the 4rth with the 5th, and so on and so forth until all you have left is the first and the very last term. But in THIS series, the 2nd cancels our the 5th, the 4th with the 7th, etc. And I can't seem to work out what would remain.. Can someone please help me?

    BTW I know the answer to be 11/30, in case that helps at all.

    Thanks in advance!
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2
    This line of thinking is the right approach. Why do you care that the terms cancelling out are further apart? You need to be a little careful, but the same ideas apply.

    The 2nd cancels the 5th, the 4th cancels the 7th, etc. So what terms do you have left? Clearly you have the 1st and 3rd term (assuming n > 1), but what happens at the end of the sum? What terms are still left? If you cannot figure this out generally right now, just try for n = 2, n =3, n =4, n= 5 and note what terms are not cancelled out. You should be able to spot a pattern.
     
  4. Feb 15, 2012 #3
    So the first and third terms will cancel out.. but Im confused about what happens as I near the end. Would it be correct to say that the sum is

    s=[itex]\frac{1}{5}[/itex]+[itex]\frac{1}{6}[/itex]+[itex]\frac{1}{n+5}[/itex]+[itex]\frac{1}{n+6}[/itex]?

    Because if so, I know how to solve it from there.
     
  5. Feb 15, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you set [tex] f(n) = \frac{1}{2n+8}, [/tex]
    you have a sum of terms [itex] t(n) = f(n) - f(n+2). [/itex] Write out a few terms to see what happens.

    RGV
     
  6. Feb 15, 2012 #5
    Try inserting n=1,n=2,n=3 and see if you expression is correct. No need to ask, just try it out.

    Anyway maybe an easier way to arrange the sum rather than the intermingled formula you came up with is to first write up the first sum and then minus the second sum:

    \begin{align*}
    \sum_{k=1}^n \frac{1}{n+4} - \sum_{k=1}^n \frac{1}{n+6} \\
    = \frac{1}{5}+\frac{1}{6}&+\frac{1}{7}+\frac{1}{8}+ \cdots + \frac{1}{n+4} \\
    &-\frac{1}{7}-\frac{1}{8}-\cdots - \frac{1}{n+4} -\frac{1}{n+5} -\frac{1}{n+6}
    \end{align*}
     
  7. Feb 15, 2012 #6
    I ended up getting the right answer, but that method you just showed me in this quote can be really useful to visualizing it in an easier way. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook