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Test the Series for Convergence or Divergence

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data
    ##\sum _{n=1}^{\infty }\left[ \left( -1\right) ^{n}\right] \dfrac {\sqrt {n}} {1+2\sqrt {n}}##


    2. Relevant equations
    Alternating Series test, Absolute convergence theorem, p-series, and test for divergence.

    3. The attempt at a solution
    The alternating series test tells us that the limit of the term An or the absolute value of the sum as it goes to infinity has to be zero and that the consecutive values of An decreasing for the series to converge.
    I found that using the first part of the test:
    lim n-->infinity of the An..
    ##\lim _{n\rightarrow \infty }\dfrac {1} {\dfrac {1} {\sqrt {n}}+2}##

    Then, I tried to use the p-series concepts(p=1/2) to conclude that 1/(n^1/2) diverges. But, I am unsure what that means for the series as a whole.

    Also, if a term diverges on the denominator. Does in necessarily make the numerator zero?
    Any help?
    Thanks
     
    Last edited: May 11, 2012
  2. jcsd
  3. May 11, 2012 #2

    sharks

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    Gold Member

    [tex]\lim _{n\rightarrow \infty } a_n =\lim _{n\rightarrow \infty }\dfrac {1} {\dfrac {1} {\sqrt {n}}+2}=\frac{1}{2}[/tex]
    The limit of the absolute value is not zero. Use the nth-term test for divergence to deduce your conclusion.
     
  4. May 11, 2012 #3
    That is the point. I don't know what 1/(n^1/2) means. All I know is that it diverges. Since, you mentioned the test of divergence probably the series diverges.
     
  5. May 11, 2012 #4

    sharks

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    Gold Member

    You cannot use a p-series test on a term within a limit. That's probably the most extreme error you could make. Like the name suggests, the p-series applies directly to series only - not to limits.

    Like i just said, you cannot deduce anything based on part of the limit. You need to consider the result of the limit as a whole. Based on the latter's value, then you can expect to make possible conclusions.
     
  6. May 11, 2012 #5
    Ok. Thank you. It is one of my common mistakes.
    Then, you are saying that 1/(n^1/2) goes to zero as n goes to infinity. It seems ok, but at first counter intuitive. I'm mistaking series with limit terms. Like you said.
     
  7. May 11, 2012 #6

    Ray Vickson

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    Homework Helper

    Look at the series
    [tex] \sum_{n=1}^\infty (-1)^n \left[ \frac{2\sqrt{n}}{1 + 2\sqrt{n}} - 1\right] =
    - \sum_{n=1}^\infty (-1)^n \frac{1}{1 + 2\sqrt{n}}. [/tex] The series
    [tex] \sum_{n=1}^\infty (-1)^n \frac{1}{1 + 2\sqrt{n}} [/tex] is convergent (by the alternating series test) so has a finite value, D. That means that for large [itex]N[/itex] we have
    [tex] \sum_{n=1}^N (-1)^n \left[ \frac{2\sqrt{n}}{1 + 2\sqrt{n}}\right] = \sum_{n=1}^N (-1)^n - D + \delta_N,[/tex] where [itex] \delta_N \rightarrow 0[/itex] as [itex] N \rightarrow \infty.[/itex] Since the series [itex] \sum (-1)^n = 1 -1 +1 -1 + \cdots + (-1)^N[/itex] does not have a finite limit as [itex] N \rightarrow \infty,[/itex], the other series does not, either.

    RGV
     
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