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Sequence proof (Hardish)

  1. May 22, 2012 #1
    http://dl.dropbox.com/u/33103477/summands.png [Broken]

    Even with the hint, I'm confused on what to use on this ? Any idea's ?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 22, 2012 #2

    sharks

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    The summand to compare with is most probably: 1/n
     
  4. May 22, 2012 #3
    Forget the hint for a second. It's a tad bit confusing (though you will see what they mean eventually).

    Can you find an upper bound for each of the ##x_n##? A lower bound?

    If that's not enough to get you started ... can you establish a ... relation ... between each of the "terms" of the ##x_n##?
     
  5. May 22, 2012 #4

    thrill3rnit3

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    [itex]\frac{1}{n+1}[/itex] = [itex]\frac{1}{\sqrt{(n+1)^2}}[/itex] = [itex]\frac{1}{\sqrt{n^2+2n+1}}[/itex] < [itex]\frac{1}{\sqrt{n^2+2n}}[/itex] < [itex]\frac{1}{\sqrt{n^2}}[/itex] = [itex]\frac{1}{n}[/itex]

    In short, we've sandwiched [itex]\frac{1}{\sqrt{n^2+2n}}[/itex] between [itex]\frac{1}{n+1}[/itex] and [itex]\frac{1}{n}[/itex]
     
  6. May 23, 2012 #5
    OK. Not exactly what I was trying to get you to do, but we might be able to make it work.

    Can you get similar bounds for the other terms? The ##\frac{1}{\sqrt{n^2+2k}}## for ##1\leq k<n##?

    While you're doing that, go ahead and figure out how many terms there are in ##x_n##.
     
  7. May 23, 2012 #6
    Doesn't the sequence diverge?

    note: n^2 + 2n ≤ 4n^2

    so 1/sqrt(n^2+ 2n) ≥ 1/sqrt(4n^2) = 1/(2n). But the sum of 1/(2n) diverges.
     
  8. May 23, 2012 #7
    The question clearly states that it converges to 1.
    So everything said till now makes no sense.
     
  9. May 23, 2012 #8
    edit: see whats going on
     
    Last edited: May 23, 2012
  10. May 23, 2012 #9
    Seconded. The series diverges. You can see it goes on increasing(>2) if you try with n>9, and the proof is right here.

    Edit : I just realized my mistake. I took into consideration that the general term is

    [itex]T_n = \frac{1}{\sqrt{n^{2}+2n}}[/itex]

    But the way the series is made, it will not have this as the general term. Gopher's method would work out best.
     
    Last edited: May 23, 2012
  11. May 23, 2012 #10
    I think that best solution (for me) is @gopher_p solution.

    You know that:

    [itex]\frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\le \frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{n^2+2}} = n\cdot \frac{1}{\sqrt{n^2+2}} =\frac{n}{\sqrt{n^2+2}}[/itex]

    On the other hand

    [itex]\frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\ge \frac{1}{\sqrt{n^2+2n}} + \frac{1}{\sqrt{n^2+2n}} + \cdots + \frac{1}{\sqrt{n^2+2n}} = n\cdot \frac{1}{\sqrt{n^2+2n}} =\frac{n}{\sqrt{n^2+2n}}[/itex]

    So, you have

    [itex]\frac{n}{\sqrt{n^2+2n}} \le \frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\le \frac{n}{\sqrt{n^2+2}}[/itex]
     
  12. May 23, 2012 #11
    Guy's this is a past exam paper question and it has come up twice. If it was a mistake it wouldn't have come up two times in exactly the same form.

    On a side note this has come up in the sequences portion of the paper(we have separate sections for different topics on the paper's easier bit), dunno if that's a hint.
     
  13. May 23, 2012 #12
    The question is correct, there is no mistake. Karamata almost gave you the solution, all you need to see is how the sandwich limits behave when n -> infinity.
     
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