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Homework Help: Convergence of Alternating Series

  1. Jan 26, 2010 #1
    I need to test the series

    ∑_(n=1)^∞〖(-1)^(n+1) 1/(n+2x^2 )〗 x in R

    for absolute, pointwise and uniform convergence.

    I am unsure about absolute convergence. I thought about using the comparison test and the idea that 1/n+2x^2 > 1/n and so diverges. Is this correct?

    I'm not sure how to go about pointwise convergence, but if shown, then I am sure I would be able to prove if it is uniformly convergent or not.

    Thanks in advance.
     
  2. jcsd
  3. Jan 26, 2010 #2

    Mark44

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    No. 1/(n + 2x^2) < 1/n. When the denominator is larger, the overall fraction is smaller. Since the inequality goes the wrong way, you can't conclude that the series diverges.
     
  4. Jan 26, 2010 #3
    Stupid me!

    I'm at a loss at how to proceed for absolute convergence now. Any ideas?
     
  5. Jan 26, 2010 #4
    If I write the series as [tex]\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n + C}[/tex] where [tex]C = 2x^2[/tex] is a constant, does that give you any ideas for what to compare it to?

    Remember that convergence or divergence is determined by the tail of the series, and that tail can start as far down as you like.
     
  6. Jan 26, 2010 #5
    isnt absolute convergence simply the sum of the absolute values of each term. Or equal to that -1 changing to a 1? And in that case you have a function with the denominator increasing by 1 each time. Which does not converge.
     
  7. Jan 26, 2010 #6
    Your comment would suggest comparing it with the Harmonic Series, which diverges.

    I have used the Ratio Test and claimed that |a(n+1)/a(n)|<1 for all x in R, as n tends to infinity. Hence converges absolutely.

    Is this correct now?
     
  8. Jan 26, 2010 #7
    Actually....I think this is wrong since it eventually equals 1, so the test is inconclusive.

    This is driving me MAD!!!!!!!!!!!!!!
     
  9. Jan 26, 2010 #8

    vela

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    Isn't there a test where you examine the limit of the ratio of the terms of a known converging or diverging series and the series in question?
     
  10. Jan 26, 2010 #9
    Here is a hint:

    Using Ystael's notation,

    [tex]
    \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n + C} = \sum_{n=C+1}^\infty \frac{(-1)^{n+1 - C}}{n}
    [/tex]
     
  11. Jan 26, 2010 #10
    The Alternating Series Test reveals that {(-1)^n+1}/n converges.
    Since {(-1)^n+1}/n+C < {(-1)^n+1}/n, then the series in question converges by the Comparison Test.

    However, this doesn't show absolute convergence.

    If I take the modulus of {(-1)^n+1}/n, then I get the Harmonic Series, which diverges.

    Confused with where to go with all this!
     
  12. Jan 26, 2010 #11

    vela

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    To show absolute convergence or divergence, you can use the limit comparison test.
     
  13. Jan 26, 2010 #12

    vela

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    That only works if C is an integer, which is not guaranteed because C=2x^2 and x can be any real number.
     
  14. Jan 26, 2010 #13
    I keep doing this mistake. : /
    Good call.
     
  15. Jan 26, 2010 #14
    And what is the issue with that? I mean n is to the power of 1, so therefore it must diverge for all values of x except infinity and negative infinity. right?
     
  16. Jan 26, 2010 #15
    In a previous post, somebody mentioned that you couldn't determine if the series diverged using this fact, since 1/n+C < 1/n. Wouldn't you need the inequality the other way around?
     
  17. Jan 26, 2010 #16

    vela

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  18. Jan 26, 2010 #17
    I'm so confused now I have no idea what I am trying to prove.

    Is the series divergent or absolutely convergent? If I know this, it might give me a starting point for a proof!
     
  19. Jan 26, 2010 #18
    Okay...but this doesn't apply since you need ALL positive terms.
     
  20. Jan 26, 2010 #19

    vela

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    You're testing for absolute convergence, right? So you're no longer looking at the original sequence, but the absolute value of the original sequence.

    (By the way, most of the other tests also require a positive or non-negative sequence.)
     
  21. Jan 26, 2010 #20
    Okay...So using the Limit Comparison Test I get a value of 1. Since 1/n is divergent, they both diverge, and so my original sequence is not absolutely convergent.

    Is that right?
     
  22. Jan 26, 2010 #21
    Sorry...value should be 1/C which is still greater than zero, since C is positive for all x in R.
     
  23. Jan 26, 2010 #22

    vela

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    Actually, you got the limit right the first time. As n goes to infinity, the value of C becomes negligible.

    You could have also used Ystael's suggestion to use the comparison test by lopping off the first N terms of the harmonic series, where N>C. Let [itex]a_n = 1/n[/itex]. If you form a new sequence by dropping the first N terms, you'd have [itex]b_n = a_{n+N}[/itex], and the corresponding series won't converge because you've dropped only a finite number of terms. Since N>C, [itex]b_n=1/(n+N)\le 1/(n+C)[/itex], according to the comparison test, your series will diverge as well.
     
  24. Jan 27, 2010 #23
    Thanks. Now for pointwise convergence!
     
  25. Jan 27, 2010 #24
    why should it matter that 1/n+C < 1/n?? say your C value is 100000000000000. then it still doesnt converge at any value. The First part of the series only will determine how large it is, but it is the evaluation of the series as n goes to infinity that is of importance.
     
  26. Jan 27, 2010 #25
    The proof I had in mind that the series does not converge absolutely actually used the comparison test against a modified sequence. This is an important thing to keep in your toolkit: if the sequence you want to compare against doesn't fit what you need exactly, you might be able to tweak it slightly so it does.

    Here is the proof I was thinking of: When n is sufficiently large, specifically when [tex]n > C[/tex], we have [tex]2n > n + C[/tex]. Therefore [tex]\frac1{n + C} > \frac1{2n}[/tex], so [tex]\sum_{n=1}^\infty \frac1{n + C}[/tex] diverges by comparison to the divergent series [tex]\sum_{n=1}^\infty \frac1{2n}[/tex].
     
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