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Finding if a Series is Convergent

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Does [itex]\sum[/itex][itex]\frac{k}{1+k^2}[/itex] converge?



    2. Relevant equations



    3. The attempt at a solution

    What I'm basically doing is using the ratio test. And at the end I get that the limit goes to ∞.
    Therefore it it is convergent. But I'm not sure if I'm using the right test or if an other test would be easier.
     
  2. jcsd
  3. Apr 25, 2012 #2

    LCKurtz

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    Without seeing your work, who knows where you went astray. I don't think the ratio test will help you on this problem. What other tests have you thought of trying?
     
  4. Apr 25, 2012 #3
    If you're getting ∞ as an answer, then the series diverges, not converges.
     
  5. Apr 26, 2012 #4
    Okay this is what I did. I decided not to use the ratio test anymore.

    http://img341.imageshack.us/img341/5567/calculus.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Apr 26, 2012 #5

    LCKurtz

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    What you have done is pointless. Noting that the nth term of a series goes to zero tells you nothing about convergence of the series.

    There are other tests for convergence/divergence than the ratio test. What are they?
     
    Last edited by a moderator: May 5, 2017
  7. Apr 26, 2012 #6
    I would try an Integral test. Always the best for "easy functions" cause you can't over think and make a mistake.
     
  8. Apr 26, 2012 #7
    Comparison test, integral test, p-series test, root and ratio tests, and the alternating series test.

    I'm just not sure which on to use. I know it can't be an alternating series, or p-series.
     
  9. Apr 26, 2012 #8

    sharks

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    Use the comparison test or the limit comparison test.

    Let [itex]u_k=\frac{k}{1+k^2}[/itex]
    Then, [itex]v_k[/itex] can be approximated to ...... as k tends towards infinity.
     
  10. Apr 26, 2012 #9

    Mark44

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    To elaborate on what LCKurtz said, the nth term test for divergence says something like this.
    "If ##\lim_{n \to \infty} a_n \neq 0##, the series Ʃ an diverges."

    This test can tell you only whether a series diverges. It is a very common mistake that students make when they conclude from this test that a given series converges. In other words, if ##\lim_{n \to \infty} a_n = 0##, you cannot conclude that the series converges.
     
  11. Apr 26, 2012 #10

    sharks

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    Example: [itex]\lim_{n \to \infty} \frac{1}{n} = 0[/itex] but it diverges! (it's a harmonic series).
     
  12. Apr 26, 2012 #11

    Mark44

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    Good example.

    Another example is ## \lim_{n \to \infty} \frac{1}{n^2} = 0##. Same result, but this time the series converges.
     
  13. Apr 26, 2012 #12

    sharks

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    Thank you for clarifying this common mistake, Mark44. I made that error not so long ago in my test.:redface: I blame it on stress.:rolleyes:
     
  14. Apr 26, 2012 #13

    LCKurtz

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    OK. Well I have already told you the ratio test won't work and you see it isn't an alternating series. It isn't a p series. So that leaves comparison and general comparison test and integral test. So try something.
     
  15. Apr 27, 2012 #14
    Okay, I tried the integral test. Does that answer look better now?

    Untitled.png

    Therefore, the seires is divergent by the integral test.
     
  16. Apr 27, 2012 #15

    Dick

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    Looks fine.
     
  17. Apr 27, 2012 #16
    The integral test works nicely. Though what I'd do is the Ratio Test. Nice solution! Really, really nice.
     
  18. Apr 27, 2012 #17

    LCKurtz

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    Which, as I have mentioned before, would be inconclusive.
     
  19. Apr 27, 2012 #18

    LCKurtz

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    Yes. See, all you had to do was try it. Good writeup.
     
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