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Alternating series

  1. Apr 2, 2005 #1
    I know that a series such as

    [tex]\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} [/tex]

    is divergent. Is this also the case for an alternating version of the same series, i.e.,

    [tex]\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}} [/tex]

    ?
     
  2. jcsd
  3. Apr 2, 2005 #2

    dextercioby

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    What criteria do you have for alternating series?

    Daniel.
     
  4. Apr 2, 2005 #3
    As dexter hinted to, look up the alternating series test. A stronger version is Abel's test, and even stronger is the Dirichlet test.
     
  5. Apr 2, 2005 #4
    Well, I know about the alternate series test, I am just saying that is it reliable to say that if a particular series does not converge to a certain sum, then a similar series that alternates between positive and negative also will not converge to a specific sum?
     
  6. Apr 2, 2005 #5
    well, it depends on the series. The alternating series test says that

    [tex]\sum_{n=0}^\infty (-1)^na_n, \; \mbox{with} \ a_n \geq a_{n+1} \ \forall n \geq 0[/tex]

    converges if [itex]\lim_{n \rightarrow \infty} a_n = 0[/itex].

    So in your specific example, the alternating series converges, even though the series is not absolutely convergent, because

    [tex]\frac{1}{\sqrt{n+1}} \leq \frac{1}{\sqrt{n}} \ \forall n \geq 1[/tex]

    and

    [tex] \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} = 0[/tex]

    If a series is convergent but not absolutely convergent, we call it conditionally convergent. Conditionally convergent series have some very unintuitive properties, one of which is described here:

    http://mathworld.wolfram.com/RiemannSeriesTheorem.html
     
  7. Apr 4, 2005 #6

    dextercioby

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    The sum is approximately [itex] 0.6 [/itex]...


    Daniel.
     
  8. Apr 4, 2005 #7

    saltydog

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    Regarding the MathWorld reference:

    Can anyone explain to me how to calculate these sums?

    [tex]\sum_{k=1}^{\infty}\frac{1}{4k(2k-1)}=\frac{1}{2}ln(2)[/tex]

    [tex]\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=ln(2)[/tex]
     
  9. Apr 4, 2005 #8

    dextercioby

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    The last is very easy,if u consider the Taylor series of [itex] \ln(1+x) [/itex] around zero...(there's another elegant construction,too).

    As for the first,write it like that

    [tex]S=\frac{1}{2}\sum_{k=1}^{+\infty}\frac{1}{2k(2k-1)}=\frac{1}{2}\sum_{k=1}^{+\infty} \left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\frac{1}{2}\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...\right]= \frac{1}{2}\ln 2 [/tex]

    ,where i made use of the second sum...

    Daniel.
     
    Last edited: Apr 4, 2005
  10. Apr 4, 2005 #9

    saltydog

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    Oh Jesus. I think you've already told me something like that before with another one. I'll spend some time with it. Thanks.
     
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