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Infinite series convergence

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the infinite series
    [itex]\sum_{n=0}^{\infty} (\sqrt{n^a+1}-\sqrt{n^a})[/itex]

    Converges for a>2 and diverges for x =2


    3. The attempt at a solution


    I'm reviewing series, which I studied a certain time ago and picking some questions at random, I can't solve this one.

    I tried every test I could think of, they were ALL inconclusive, I believe there's some function I must came up with to use with the comparison test, but I can't figure which one is it.
     
  2. jcsd
  3. Jan 27, 2013 #2

    LCKurtz

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    Did you try rationalizing the numerator first?
     
  4. Jan 27, 2013 #3
    Ohh, no I didn't. I didn't even think about that, but now that you mentioned it, it worked perfectly, I could apply the comparison test in the rationalized numerator and find my answer, beautiful!

    Thanks a LOT LCKurtz :D

    Ps: I really wish I had more of a "math view" that you guys in this forum frequently have, and identify when I should rationalize, when I should expand fractions, etc.
     
  5. Jan 27, 2013 #4

    Mark44

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    Since x isn't present in this problem, I assume you mean a = 2, and possibly a ≤ 2.
     
  6. Jan 27, 2013 #5
    Hey mark44!

    Yes, I apologize, I actually meant a>2 and not x>2, and yes, it's bigger than. If you do the rationalization it will become clear that you can compare the a=2 (when it becomes divergent) with the harmonic series, and a>2 will have an exponent smaller than one (p-series with p <1, that is, convergent.)
     
  7. Jan 27, 2013 #6

    Mark44

    Staff: Mentor

    You wrote a > 2 for convergence, which was OK. The other part had to do with divergence of the series, and you wrote x = 2. My correction was to change this to a = 2.
     
  8. Jan 28, 2013 #7
    Ok so I thought I figured out with the rationalization tip, but I think I only solved half of the problem with it:

    [itex]\frac{1}{\sqrt{n^a +1}+\sqrt{n^a}} < \frac{1}{\sqrt{n^a}}[/itex] For all n and a>0

    So, applying the comparison test:

    [itex]\sum_{n->0}^{\infty}\frac{1}{n^{a/2}}[/itex]

    Which converges for p>1, thus, for a>2. Great, I just provided proof that it converges for a>2.

    While it's obvious that when a=2 it becomes the harmonic series, I can't use that in the proof, because I'm comparing a smaller series with a bigger one, so I can only proof convergence, how would I go about providing proof of divergence?

    Thanks in advance.
     
  9. Jan 28, 2013 #8

    LCKurtz

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    Try making the denominator a little bigger but still keeping it "like" a harmonic series.
     
  10. Jan 28, 2013 #9
    Hey LCKurtz!
    Thanks again for all your help, I did it, but I did using the limit comparison test this time, with:
    [Itex]b_n=\frac{1}{n}[/itex]

    It worked flawlessly! Out of curiosity, though did you find another solution? I just ask because you mentioned "making it bigger".
     
  11. Jan 28, 2013 #10

    LCKurtz

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    I used the estimate$$
    \frac 1 {\sqrt{n^a+1}+\sqrt{n^a}}\ge \frac 1 {\sqrt{n^a+1}+\sqrt{n^a+1}}
    =\frac 1 {2\sqrt{n^a+1}}$$and used the limit comparison on that with ##\frac 1 n## for ##a=2##.
     
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