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Absolute or Conditional Convergence, or Divergence of Alternating Series.

  1. Jul 13, 2011 #1
    1. The problem statement, all variables and given/known data
    given an = ( -1 )(n+1) / [itex]\sqrt{n}[/itex] determine if the infinite series is Absolutely Convergent, Conditionally Convergent, or Divergent.



    2. Relevant equations

    I hope I have these theorems down correctly, please correct me if i'm wrong. If [itex]\Sigma[/itex]|an| is Convergent then [itex]\Sigma[/itex]an is Absolutely Convergent if [itex]\Sigma[/itex]|an| is Divergent and [itex]\Sigma[/itex]an is Convergent then [itex]\Sigma[/itex]an is Conditionally Convergent And if neither occur then the [itex]\Sigma[/itex]an may be Divergent.

    3. The attempt at a solution

    To show [itex]\Sigma[/itex]|an| or [itex]\Sigma[/itex]an are convergent I have to show 2 things, 1). lim[ n[itex]\rightarrow[/itex][itex]\infty[/itex] ] |an| = 0 and that either an+1 [itex]\leq[/itex] an, an+1/an [itex]\leq[/itex] 1 or d(an)/dn < 0 which determine whether or not an is Decreasing. If I can show these things I have a convergent Alternating Series, if not I might be able to try another test or I have a divergent Alternating Series.

    So I started by seeing if lim[ n [itex]\rightarrow[/itex] [itex]\infty[/itex] ] |an| = 0 so I have to figure out what |an| actually is. Again correct me if i'm wrong, it would help me out greatly here, but I believe |(-1)n+1/[itex]\sqrt{n}[/itex]| to be 1/[itex]\sqrt{n}[/itex] because |+/-(1)| = 1 there for I can determine that lim[ n [itex]\rightarrow[/itex] [itex]\infty[/itex] ] 1/[itex]\sqrt{n}[/itex] = 1/[itex]\infty[/itex] = 0 so if I have my thoughts in order I have shown lim[n [itex]\rightarrow[/itex] [itex]\infty[/itex] ] 1/[itex]\sqrt{n}[/itex] = 0 great! Next I have to show that an+1 [itex]\leq[/itex] an so I first considered [itex]\sqrt{n+1}[/itex] which[itex]\geq[/itex] [itex]\sqrt{n}[/itex] so 1/|[itex]\sqrt{n+1}[/itex]| [itex]\leq[/itex] 1/|[itex]\sqrt{n}[/itex]|
    to further solidify this in my mind or at least on paper, I wanted to show that d(|an|/dn [itex]\leq[/itex] 0 so I did just that and acquired -1/(2n3/2) [itex]\leq[/itex] 0 [itex]\forall[/itex] [itex]\Re[/itex] therefor [itex]\Sigma[/itex]an is Absolutely Convergent. The book says that it is conditionally convergent, I tried the integral test for no good reason and paired with my above results got the exact opposite of my definition for conditional convergence.

    After moving on I got the subsequent problems incorrect as well, falsly concluding that my series was Absolutely Convergent when they were Conditionally Convergent so I can't really move anywhere until I understand this, thanks in advance for your help.
     
  2. jcsd
  3. Jul 13, 2011 #2

    lanedance

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    Homework Helper

    i think you're on the right track, the series is conditionally convergent shown by applying the "alternating series test"
    -an tends to zero monotonically and changes sign
    so the series as its written it converges to a finite limit

    however as you've also correctly worked out it is not absolutely convergent as the sum of the modulus of each term diverges

    any absolutely convergent series is conditionally convergent, however the converse does not apply
     
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