1. Not finding help here? Sign up for a free 30min 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!

Ars Conjectandi Series

  1. Nov 4, 2011 #1
  2. jcsd
  3. Nov 4, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: Expand [itex]\frac 1 {n(n+1)}[/itex] into partial fractions.
     
  4. Nov 4, 2011 #3

    Mark44

    Staff: Mentor

  5. Nov 4, 2011 #4

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    1. But the difference between two divergent series is not necessarily divergent or infinite.

    2 Actually write stuff out - write out Sn - Sn+1 as series and what does it equal? and write out the term typical term 1/n(n + 1) - how else would you naturally express that?

    This should suggest something to you.
     
  6. Nov 4, 2011 #5
    Well, I realized that Sn - Sn+1 = S the first series. If I multiply that by 2 I get the sum identity inverse.

    So, he's wanting an expression for the finite sum? I suspected it was finite, but I wasn't sure.

    Terms would cancel in that difference. The first series is a telescoping series.

    It simplifies to 1 - (1/n+1) for some n.
     
    Last edited: Nov 4, 2011
  7. Nov 5, 2011 #6

    Mark44

    Staff: Mentor

    Both of the sums shown in the linked-to image are finite sums.
    You don't have the parentheses in the right places. This should be 1 - 1/(n+1).
     
  8. Nov 5, 2011 #7

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    [tex]\frac{1}{n(n + 1)} = ? [/tex]

    Edit (Actually that is what Kurz is saying).

    Sooner or later you'll kick yourself.
     
    Last edited: Nov 5, 2011
  9. Nov 5, 2011 #8
    Yeah, I have 1 - 1/(n+1) on my paper. If you take the limit as n goes to infinity it tends to 1. Since n does not go to infinity, then you have the expression for some finite n.
     
  10. Nov 5, 2011 #9

    Mark44

    Staff: Mentor

    Based on the photo you uploaded, the question has nothing to do with limits. It could be that this is asked for in the problem itself, but it looks to me like what they're asking you to do is find Sn = 1 + 1/2 + 1/3 + ... + 1/n, which I say again, is a finite sum.
     
  11. Nov 5, 2011 #10
    Sn - Sn+1 = 1 - 1/(n+1).
     
  12. Nov 5, 2011 #11

    Mark44

    Staff: Mentor

    This is incorrect - the left side is negative and the right side is close to 1 (hence positive).
     
  13. Nov 5, 2011 #12
    Are you sure about that?

    Sn - Sn+1 = (1 + 1/2 + 1/3 + ... + 1/n) - (1/2 + 1/3 + ... + 1/n + 1(n+1)) = 1 - 1/(n+1).
     
  14. Nov 5, 2011 #13
    If I write this in partial fractions I get

    1 = A(n+1) + Bn

    If I continue, I don't get a nice expression.
     
  15. Nov 5, 2011 #14

    Mark44

    Staff: Mentor

    Yes.
    The expression you have for Sn+1 is wrong because it is missing a term. Sn is a sum of n terms, while Sn+1 is a sum of n + 1 terms. Sn - Sn+1 < 0.
     
  16. Nov 5, 2011 #15

    Mark44

    Staff: Mentor

    ???

    The equation above is an identity that must be true for all n. That means that the polynomial on the left has to be identically equal to the one on the right.

    Grouping by powers of n gives
    1 = (A + B)n + A

    More suggestively, this is
    0n + 1 = (A + B)n + A
     
  17. Nov 5, 2011 #16
    I'm interpreting it differently. Why is it not the following?

    sum of (1/k) from k = 1 to k = n

    sum of (1/k+1) from k = 1 to k = n
     
  18. Nov 5, 2011 #17
    I haven't worked partial fractions in a while. I reworked and got 1 = A and -1 = B.

    According to what you're saying,

    Sn - Sn+1 = (1 + 1/2 + 1/3 + ... + 1/n) - (1 + 1/2 + 1/3 + ... + 1/n + 1(n+1)) = -1/(n+1).
     
    Last edited: Nov 5, 2011
  19. Nov 5, 2011 #18

    Mark44

    Staff: Mentor

    Yes, that's better. Now, do you understand how all this ties in to the real problem?
    Namely, finding the sum
    [tex]\frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

    The whole business of partial fraction decomposition is intended to help you rewrite the individual terms in the sum above.
     
  20. Nov 5, 2011 #19
    Maybe.

    S = 2Sn - Sn+1
     
  21. Nov 5, 2011 #20

    Mark44

    Staff: Mentor

    How are you getting this?

    I worked this problem using the suggested hint and have something completely different for S.

    One thing that bothers about the problem statement is their confusing use of S and Sn to represent unrelated things. For example, in the problem it is given that

    [tex]S = \frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

    Later on, they have Sn = 1 + 1/2 + 1/3 + ... + 1/n. For this latter sum, they should have used a different letter altogether, maybe Hn.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook