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Determine whether the series is convergent

  1. Oct 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series [tex]\sum_{n=2}^{\infty}a_n[/tex] is absolutely,conditionally convergent or divergent

    3. The attempt at a solution
    from Abel's test.[tex]c_n=\frac{(-1)^n}{\sqrt{n}}[/tex]is convergent.and

    [tex]b_n=(\frac{2n}{n+1})^\pi}=\frac{2^{\pi}}{(1+\frac{1}{n})^{\pi}}=\frac{2^{\pi}}{1+\frac{\pi}{n}+o(\frac{1}{n^2})}[/tex].Which has limit [tex]2^{\pi}[/tex].So a_n is convergent.


    I don't know exactly but it seems to me that the last equation is divergent.So a_n is conditionally convergent.
  2. jcsd
  3. Oct 26, 2007 #2
    Sorry,I wrongly had typed the series .
  4. Oct 26, 2007 #3


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    I don't think you can apply Abel's test here, since the basic series you are dealing with, 1/sqrt(n) is of the form 1/n^p and doesn't converge by an integral test. I would look for a proof using the alternating series test.
  5. Oct 27, 2007 #4
    The reason why I used Abel's Test is:

    Abel's Test

    Given two sequences {[tex]a_n[/tex]} and{[tex]b_n[/tex]},suppose that
    1.The series
    [tex]\sum_{n=1}^{\infty}a_n[/tex] is convergent.
    2.The sequence [tex]b_n[/tex] monotonically converges to some number L
    Then the series
    [tex]\sum_{n=1}^{\infty}a_nb_n[/tex] is convergent.

    So I choosed first series [tex]c_n=\frac{(-1)^n}{\sqrt{n}}[/tex] as convergent.And [tex]b_n[/tex] with limit [tex]2^{\pi}[/tex].
  6. Oct 27, 2007 #5
    Let [tex]a_n = \frac{(-1)^n}{\sqrt{n}}[/tex] then [tex]\sum_{n=1}^{\infty} a_n [/tex] converges. Let [tex]b_n = \left( \frac{2n}{n+1} \right)^n[/tex] then [tex]b_n \leq b_{n+1}[/tex] and [tex]\lim b_n \not = \infty[/tex]. This means [tex] \sum_{n=1}^{\infty} a_nb_n[/tex] converges.
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