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Proving this series converges or diverges

  1. Jul 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove [itex] y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} [/itex] where [itex]n\in\mathbb{N}[/itex] is convergent or divergent.


    2. Relevant equations



    3. The attempt at a solution
    I am stuck on this. I think its divergent but I am having trouble proving that it is divergent. I tried to break up the sequence using partial sums but I don't see a pattern. What I also don't get is what [itex]y_1[/itex] is equal too. Is it [itex]y_1=\frac{1}{1+1}+\frac{1}{1+2}+...+\frac{1}{2*1} [/itex] or is it equal to [itex]y_1=\frac{1}{1+1} [/itex]? I've been stuck for quite a while.
     
    Last edited: Jul 6, 2013
  2. jcsd
  3. Jul 6, 2013 #2
    From what I understand, it looks like this:
    $$y_n=\sum_{k=n+1}^{2n}\frac{1}{k}.$$
    That help?
     
  4. Jul 6, 2013 #3
    Yes but how would I go about proving it is divergent. I was thinking of using [itex] \frac{1}{n^2}+...+\frac{1}{n^2}[/itex] since [itex] \frac{1}{n+1}+...+\frac{1}{2n}>\frac{1}{n^2}+...+\frac{1}{n^2} [/itex]?{Edit*I don't think it works}. I was thinking of using [itex] \frac{1}{n+1}+...+\frac{1}{n+1}[/itex]. I don't think the way you wrote the series works. I think the limits have to be adjusted.
     
    Last edited: Jul 6, 2013
  5. Jul 6, 2013 #4

    HallsofIvy

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    No, that doesn't work. Since the denominators are all "of order n", it would be better to compare the terms to 1/n (or your suggestion 1/(n+1)). For all i between 1 and n, [itex]1/(n+i)\ge 1/(2n)[/itex]
     
  6. Jul 6, 2013 #5
    I...don't think so. Check again. I'm pretty sure my expression for the summation is correct.

    I'd use a test of convergence. Are you familiar with any?
     
  7. Jul 6, 2013 #6
    Well this what I get so far that [itex] \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1} [/itex] Hence its divergent. I think it works since [itex] \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}[/itex] ≤ [itex] \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} [/itex]
     
    Last edited: Jul 6, 2013
  8. Jul 6, 2013 #7

    lurflurf

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    This is a little bit subtle

    $$\frac{1}{2}=\frac{n}{2n}<\sum_{k=1}^n \frac{1}{n+k}<\frac{n}{n+1}<1$$

    hint:think log(n)

    hint2:
    if

    $$H_n=\sum_{k=1}^n \frac{1}{k}$$

    then

    $$\sum_{k=1}^n \frac{1}{n+k}=H_{2n}-H_{n}$$
     
    Last edited: Jul 6, 2013
  9. Jul 6, 2013 #8
    so is the way I have it not correct then?
     
  10. Jul 6, 2013 #9

    lurflurf

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    ^The way you wrote the sum is right, but you cannot conclude it diverges from what you have done.
     
  11. Jul 6, 2013 #10

    Dick

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    If you know what a Riemann sum is, write one for the integral of the function 1/x from x=1 to x=2.
     
  12. Jul 6, 2013 #11
    Ok I will try other methods. I would think there would be a much simpler way of doing this because Bartle (book im using) presents convergence and divergence of series before he introduces integration and differentiation.
     
  13. Jul 6, 2013 #12

    Dick

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    Fair enough. You can also show a sequence converges if you can show it's monotone and bounded. Can you show ##y_{n+1}-y_{n}## is positive so it's monotone increasing and, as lurflurf indicated, bounded above?
     
  14. Jul 6, 2013 #13
    So [itex]y_{n+1}= \frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+1}[/itex] and [itex] y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}[/itex]. So [itex] y_{n+1}-y_n= \frac{-1}{n+1}+\frac{1}{2n+1}[/itex]. I think that is how it supposed to go.
     
  15. Jul 6, 2013 #14

    Dick

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    Right idea, but the last term in ##y_{n+1}## is ##\frac{1}{2(n+1)}##, not ##\frac{1}{2n+1}##. Do it a little more carefully. It might be good to write out the n=1, n=2, n=3 cases explicitly, then do some differences and compare with what you think the answer is symbolically. That's what I do.
     
    Last edited: Jul 6, 2013
  16. Jul 6, 2013 #15
    Oh yeah I forgot. [itex]y_{n+1}= \frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+5}[/itex] and [itex] y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}[/itex] So [itex] y_{n+1}-y_n=\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}[/itex] which seems to be decreasing that is [itex] y_{n+1}\leq y_n[/itex]
     
  17. Jul 6, 2013 #16

    Dick

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    Ok, so why do you think [itex]\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}[/itex] is negative? I don't. Just realized 2n+5 has got to be a typo, since your difference is correct.
     
    Last edited: Jul 6, 2013
  18. Jul 6, 2013 #17
    Oh yes its increasing. Now once I show that [itex] \frac{-1}{n+1} \leq \frac{1}{2n+1}+\frac{1}{2n+2}[/itex]. Now I think I'm going to have to use a epsilon proof to show [itex] y_n[/itex] its convergent which I need some help with.
     
  19. Jul 6, 2013 #18

    Dick

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    That's a little garbled too. My whole point is that there is likely a theorem in Bartle that if a sequence is monotone increasing and bounded above, then it converges. Use the theorem. No need for epsilons if you've got the theorem.
     
    Last edited: Jul 6, 2013
  20. Jul 6, 2013 #19
    It says 3.3.2 Monotone Convergence Theorem A monotone, sequence of real numbers is
    convergent if and only if it is bounded. Further: (a) If X = (xn) is a bounded increasing sequence, then [itex] lim(x_n) = sup\{x_n : n ε N\} [/itex]. I some how have to show [itex]y_n[/itex] is bounded and all I know its increasing so it is monotone.
     
  21. Jul 6, 2013 #20

    Dick

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    lurflurf already hinted you that it's bounded. You are summing n numbers and they are all less than 1/n, so? Doesn't that mean it's bounded? Are you clear on the monotone bit? Why is ##\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}## positive?
     
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