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Homework Help: Ars Conjectandi Series

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

    LCKurtz

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

    Mark44

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  5. Nov 4, 2011 #4

    epenguin

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    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

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    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

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    [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

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    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

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    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

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    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

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    ???

    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

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    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

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    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.
     
  22. Nov 5, 2011 #21
    Well, maybe that's my fault. I'm calling that S. The problem just gives the series. It doesn't call it S or anything.

    Here's how I got that expression up there. You should be able enlarge it.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111105_142452.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  23. Nov 5, 2011 #22

    Mark44

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    Then lets' not give names to 1 + 1/2 + ... + 1/n and 1 + 1/2 + ... + 1/n + 1/(n+1). And let's reserve Sn to mean the first n terms of the original sum. IOW,

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

    As I see it, you aren't making the connection between the partial fractions business and the terms in this series. The partial fractions decomposition you did says that 1/(n(n+1)) = 1/n - 1/(n + 1).

    Apply this formula to each term in the sum above to get a formula for Sn
     
  24. Nov 5, 2011 #23
    Your Sn is different than their first series. You can express your Sn as sum of n[(1/n) - 1/(n+1)] from 1 to n.
     
  25. Nov 5, 2011 #24

    LCKurtz

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    This thread has gotten so bollixed up I can't tell whether Shackleford has solved the problem or not. The original sum was

    [tex]\sum_{k=1}^n \frac 1 {k(k+1)} = \sum_{k=1}^n \left(\frac 1 {k}-\frac 1 {1+k}\right)[/tex]

    Just write that out and cancel what you can and it's done.
     
  26. Nov 5, 2011 #25
    That's the sum I interpreted initially.

    You get 1 - 1/(n+1), right?

    And using partial fractions you show that the above equals the desired series.
     
    Last edited: Nov 5, 2011
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