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Looks like the Harmonic series

  1. Oct 21, 2011 #1
    does let z[itex]_{n}[/itex]=

    [itex]\frac{1}{n+1}[/itex]+[itex]\frac{1}{n+2}[/itex]+[itex]\frac{1}{n+3}[/itex]+.....+[itex]\frac{1}{2n}[/itex]

    does z[itex]_{n}[/itex] converge or diverge..


    I want to say it diverges because it looks like the Harmonic series
     
  2. jcsd
  3. Oct 21, 2011 #2
    Re: Sequences

    yep I am positive
     
  4. Oct 21, 2011 #3

    Mark44

    Staff: Mentor

    Re: Sequences

    Each term in the sum above is >= 1/(2n), which is the smallest term, and there are n terms. This means that zn >= n/(2n) = 1/2.

    Also, each term in the sum is <= 1/(n + 1), which is the largest term, and there are still n terms. This means that zn <= n/(n + 1).

    One of the tests you have learned is applicable here.
     
    Last edited: Oct 21, 2011
  5. Oct 21, 2011 #4
    Re: Sequences

    does that mean it converges?
     
  6. Oct 21, 2011 #5

    Mark44

    Staff: Mentor

    Re: Sequences

    Write down a few terms in your sequence.
    z1 = 1/2
    z2 = 1/3 + 1/4
    z3 = 1/4 + 1/5 + 1/6

    What can you say about
    [tex]\lim_{n \to \infty} z_n~?[/tex]

    Note that I added some to my previous response.
     
  7. Oct 21, 2011 #6
    Re: Sequences

    [tex]
    \begin{array}{rcl}
    z_{n + 1} - z_{n} & = & \left( \frac{1}{n + 2} + \ldots + \frac{1}{2 n} + \frac{1}{2 n + 1} + \frac{1}{2 n + 2} \right) \\

    & - & \left(\frac{1}{n + 1} + \frac{1}{n + 2} + \ldots + \frac{1}{2 n} \right) \\

    & = & \frac{1}{2 n + 1} + \frac{1}{2 n + 2} - \frac{1}{n + 1} \\

    & = & \frac{1}{2 n + 1}- \frac{1}{2 n + 2} > 0
    \end{array}
    [/tex]

    Post #4 holds the other clue.
     
    Last edited: Oct 21, 2011
  8. Oct 21, 2011 #7
    Re: Sequences

    as n goes to infinity isnt the limit 1? because of zn <= n/(n + 1).
     
  9. Oct 21, 2011 #8

    Mark44

    Staff: Mentor

    Re: Sequences

    About all you can say is that 1/2 <= zn <= 1, but that doesn't mean that the limit has to be either of the endpoints.
     
  10. Oct 21, 2011 #9
    Re: Sequences

    since its bounded and increasing it converges doesnt
     
  11. Oct 21, 2011 #10
    Re: Sequences

    no. look at, for example:
    [tex]
    z_{2 n} = \left( \frac{1}{2n + 1} + \ldots + \frac{1}{3 n} \right) + \left(\frac{1}{3 n + 1} + \ldots + \frac{1}{4 n}\right) \le \frac{n}{2 n + 1} + \frac{n}{3 n + 1} = \frac{n (5 n + 2)}{(2 n + 1)(3 n + 1)}
    [/tex]
    [tex]
    z_{2 n} \le \frac{5 n^2 + 2 n}{6 n^2 + 5 n + 1} = \frac{1}{\frac{6 n^2 + 5 n + 1}{5 n^2 + 2 n}} = \frac{1}{\frac{6}{5} + \frac{\frac{13 n}{5} + 1}{n (5 n + 2)}} \le \frac{1}{\frac{6}{5}} = \frac{5}{6}
    [/tex]

    Thus, there is a subsequence that is definitely not within a neighborhood [itex]\epsilon = 1/5[/itex] of 1. So, 1 is not a limit of the sequence.

    yes, but it's actually decreasing.
     
    Last edited: Oct 21, 2011
  12. Oct 21, 2011 #11
    Re: Sequences

    ok since its monotonic decreasing and bounded it is convergent
     
  13. Oct 21, 2011 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Sequences

    It is NOT decreasing, it's increasing. I'd suggest to Punkyc7 not to believe everything everyone tells you.
     
  14. Oct 21, 2011 #13
    Re: Sequences

    yes, and the limit is somewhere between 1/2 and 5/6. Actually, the exact sum is expressible via elementary functions. Do you know what it is?
     
  15. Oct 21, 2011 #14
    Re: Sequences

    read post #6.
     
  16. Oct 21, 2011 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Sequences

    I did. And I have no idea what you are doing. z_1=1/2, z_2=7/12, z_3=37/60. Now you read post #6 and tell me what's wrong with it.
     
  17. Oct 21, 2011 #16
    Re: Sequences

    there's an extra term -1/n that does not belong there.
     
  18. Oct 21, 2011 #17

    Dick

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    Science Advisor
    Homework Helper

    Re: Sequences

    Perhaps. I didn't read it THAT carefully. But I do know z_n is an increasingly accurate estimate to log(2). And all of the z_n are underestimates. Think of it as like an integral estimate.
     
  19. Oct 21, 2011 #18
    Re: Sequences

    I corrected it.

    Look at my post #13.
     
  20. Oct 21, 2011 #19

    Dick

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    Science Advisor
    Homework Helper

    Re: Sequences

    Ok, so you know what it is. That's good. I'd still suggest to Punkyc7 to think more about the problem. And maybe for you to maybe provide less details until Punkyc7 does that.
     
  21. Oct 22, 2011 #20

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Sequences

    The original sum is a sum of positive terms. If we are always adding a positive number, the sum must be increasing.
     
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