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Sum of a Miscelleneous series.

  1. Oct 22, 2011 #1

    AGNuke

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    If [itex]f(r) = 1 + \frac{1}{2} + \frac{1}{3}+... \frac{1}{r}[/itex] and [itex]f(0) = 0[/itex], then find [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex]

    I tried the following steps but stuck at some point.

    [itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r)[/itex]

    [itex]= \sum_{r=1}^{n}rf(r)+(r+1)[f(r+1)-\frac{1}{r+1}][/itex]

    [itex]= \sum_{r=1}^{n}rf(r)+(r+1)f(r+1) - \sum_{r=1}^{n}1[/itex]

    The real problem is to find [itex]\sum_{r=1}^{n}rf(r)+(r+1)f(r+1)[/itex]

    Can someone please help me beyond this point by solving the above steps?

    For verification, the Solution is [itex](n+1)^{2}f(n+1)-\frac{n^2+3n+2}{2}[/itex]

    Thanks in Advanced!
     
    Last edited: Oct 22, 2011
  2. jcsd
  3. Oct 22, 2011 #2

    Mark44

    Staff: Mentor

    Then [itex]\sum_{r=1}^{n}(2r+1)f(r)[/itex] does what?
    Don't use [itex]\Rightarrow[/itex] here. Connect expressions that are equal with =.
     
  4. Oct 22, 2011 #3

    AGNuke

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    Fixed

    Fixed
     
  5. Oct 23, 2011 #4

    AGNuke

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    Anyone even with the slightest idea what to do next? I don't have much time left to submit the solution and I am also working on the solution simultaneously.

    Thanks!
     
  6. Oct 23, 2011 #5

    I like Serena

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    Hi AGNuke! :smile:

    I do not know how to derive your solution, but you can prove it with full induction.
    Would full induction satisfy the requirements?
     
  7. Oct 23, 2011 #6

    AGNuke

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    It's a problem of Summation for a product of two functions of a natural number. One is quite easy but the given f(r) is quite complex to sum.

    I am like in 11th grade or something, so I really don't know much about advanced mathematics.
     
  8. Oct 23, 2011 #7

    I like Serena

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    I'd classify any derivation or proof as advanced mathematics.

    Without it I can only recommend trial and error.
    If you have multiple answers you can verify the answers by trying them for n=1 and n=2.
    And if that doesn't leave one answer, try n=3.
     
  9. Oct 23, 2011 #8

    AGNuke

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    That's what we call objective approach. The answer is satisfying n=1. But will it not be good if I can do the same with the proper subjective approach?
     
  10. Oct 23, 2011 #9

    I like Serena

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    I'm not sure what you mean by subjective approach, but I do not see another approach that you can complete within 3 minutes.
     
  11. Oct 23, 2011 #10

    AGNuke

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    Subjective approach is pure mathematics + Some tricks based on assumptions and previous knowledge to cut down the time.

    I know this question is ultra level, that's why we weren't able to solve it. But since we've got plenty of time after the paper, I was considering that I can get a proper subjective approach on papers.
     
  12. Oct 23, 2011 #11

    I like Serena

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    Well, I guess I can give you an estimation method.

    We have these things called the order of a sequence or series.
    For instance 2n+1 has order n, and n^2 + 3 has order n^2.
    The series [itex]\sum\limits_r 2r+1[/itex] has order n^2, since it's an arithmetic progression.

    Since f(r) is a diverging series, but diverges slower than r, it has order n.

    With these you can determine the order of your original expression and check what the order of your answers is.
     
  13. Oct 23, 2011 #12

    AGNuke

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    Ok... Thanks for this method, but anyways, I got my answer after an hour or so of paperwork. I'll be posting it shortly to see where could I cut through steps and time. :biggrin:
     
  14. Oct 24, 2011 #13

    AGNuke

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    Here's my attempt at the question. Got my answer but still I feel it is quite unorthodox. I also want to see where can I cut short some steps.

    gif.latex?f(r)=%201+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{r};%20\sum_{r=1}^{n}(2r+1)f(r).gif

    gif.latex?=\sum%20[r+(r+1)]f(r).gif

    gif.latex?=\sum%20rf(r)+\sum%20(r+1)f(r).gif

    gif.latex?=\sum%20rf(r)+\sum%20(r+1)\left%20[%20f(r+1)-\frac{1}{(r+1)}%20\right%20].gif

    gif.latex?=\sum%20rf(r)+\sum%20(r+1)f(r+1)-\sum%201.gif

    We know that gif.latex?\sum_{r=1}^{n}1=n.gif and expanding Ʃrf(r)+Ʃ(r+1)f(r+1), we get

    gif.latex?\dpi{120}%20\left%20[%201f(1)+2f(2)+3f(3)+...gif gif.latex?\dpi{120}%20\left%20[2f(2)+3f(3)+...gif

    gif.latex?\dpi{120}%20=\left%20[%201f(1)+2f(2)+3f(3)+...gif gif.latex?\dpi{120}%20\left%20[1f(1)+2f(2)+3f(3)+...gif

    gif.gif

    As we can see, 1f(1) is simply equals to 1.

    Expanding rf(r) so as to find something more useful.

    gif.gif gif.gif gif.gif gif.gif gif.gif gif.latex?\dpi{120}%20+...+\left%20(\frac{n}{1}+\frac{n}{2}+\frac{n}{3}+%20\frac{n}{4}+...gif

    If we group them properly, what we see, we see some individual series with a common denominator, and what's more, they are in AP! And taking out their sum is no issue.

    gif.gif gif.latex?\dpi{120}%20=\left%20(%20\frac{1}{1}+\frac{2}{1}+\frac{3}{1}+\frac{4}{1}+...gif gif.latex?\dpi{120}%20+\left%20(%20\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...gif gif.latex?\dpi{120}%20+\left%20(%20\frac{3}{3}+\frac{4}{3}+...gif gif.latex?\dpi{120}%20+...gif

    Sum of AP in the first bracket is gif.gif

    Sum of AP in the second bracket is gif.gif

    Sum of AP in the third bracket is gif.gif

    and if we continue to follow the trend, we get the sum of the last bracket as gif.gif

    Here's my mischievous trick : If I consider another variable term k, which is the order of the bracket : first, second, third,..., nth bracket as k = 1, 2, 3,..., n. This forms a series of sum of sum of different APs, and the term of the main series is given by:

    gif.gif

    By doing this dirty work, I converted the summation of seemingly-impossible-to-sum series into a nicer series. Hence,

    gif.gif

    Using this equation into long forgotten step:

    gif.gif

    Expanding our new summation function, we get

    gif.gif
    (Sorry for eating out intermediate step, was too lazy to type that out)

    By observation, we see gif.gif ; gif.gif and gif.gif

    By further manipulation,

    gif.gif

    Therefore, the next step is

    gif.gif

    Grouping common terms:

    gif.gif

    gif.gif

    Tada! Here comes the answer but at this cost... The thing I am asking for is How to cut short through the procedure via a better trick or a concept (not too high level, I'm only in 11th grade)

    Hoping for a reply soon!
     
    Last edited: Oct 24, 2011
  15. Oct 24, 2011 #14

    I like Serena

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    Congratulations! Looks good!! :smile:

    I see you've already taken the time to make it as readable as possible.
    The bad news: I see no possible shortcuts.
     
  16. Oct 24, 2011 #15

    AGNuke

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    What's the use of a solution if even I can't read it.

    Sorry to hear that you can't see any possible shortcuts. Anybody else?
     
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