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Convergence of partial sums

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Give an example of a sequence [tex](a_n)[/tex] so that [tex]lim_{n\rightarrow\infty} \left|a_{n+1}/a_{n}\right| =1[/tex] and [tex]\sum^{\infty}_{n=1} a_{n}[/tex] converges


    2. Relevant equations

    (Maybe relevant, maybe not)
    Theorem which states:

    If [tex]\sum^{\infty}_{n=1} a_{n}[/tex] converges, then [tex]lim_{n\rightarrow\infty} a_{n} =0[/tex]

    3. The attempt at a solution

    I'm having trouble coming up with [tex]\sum^{\infty}_{n=1} a_{n}[/tex] that converges...

    Since [tex]lim_{n\rightarrow\infty} a_{n} =0[/tex] doesn't imply the convergence of [tex]\sum^{\infty}_{n=1} a_{n}[/tex] (the theorem only works the other way around), I'm not sure where to start.

    Any hint will be appreciated. Thank you.
     
  2. jcsd
  3. Oct 7, 2009 #2

    Landau

    User Avatar
    Science Advisor

    You can certainly come up with some converging sum, ignoring the other requirement for a moment?

    [tex]a_n=\frac{1}{n}[/tex] satisfies [tex]\left|\frac{a_{n+1}}{a_n}\right|\to 1[/tex], but [tex]\sum^{\infty}_{n=1} a_{n}[/tex] does not converge. Can you modify this example such that it does?
     
  4. Oct 7, 2009 #3
    Sorry, I don't think my question was clear. I'm having trouble understanding what makes certain partial sums converge/not converge.

    So yeah, [tex]a_n=\frac{1}{n}[/tex] does satisfy [tex]\left|\frac{a_{n+1}}{a_n}\right|\to 1[/tex], but why wouldn't [tex]\sum^{\infty}_{n=1} a_{n}[/tex] converge? Isn't [tex]1/n[/tex] approaching zero?
     
  5. Oct 7, 2009 #4
    Last edited: Oct 7, 2009
  6. Oct 8, 2009 #5

    Mark44

    Staff: Mentor

    The purpose of this exercise seems to be exploring the edge case of the ratio test. For the terms in an infinite series [itex]\sum a_n[/itex], you look at the limit
    [tex]\lim_{n \rightarrow \infty}\left|\frac{a_{n + 1}}{a_n}\right|~=~L[/tex]
    If L < 1, the series converges absolutely.
    If L > 1, the series diverges.
    If L = 1, or no limit exists, the test is inconclusive.

    This problem seems to be about that third possibility, where L = 1.
     
  7. Oct 8, 2009 #6
    Okay, I think I understand it... so if L=1, it's possible for [tex]\sum a_{n}[/tex] to either converge or diverge.

    So, for example, when [tex]a_{n}[/tex] is an alternating harmonic series (from the wiki article above):
    1c17b395c60e0e044a7235fc4dcc6915.png

    then [tex]\sum a_{n}[/tex] would converge and also have L=1.

    On the other hand, if [tex]a_{n}[/tex] is something like [tex]a_{1}=1, a_{n+1}=a_{n}[/tex] so that [tex](a_{n}) = (1, 1, 1, 1, 1, ...)[/tex]

    then [tex]\sum a_{n}[/tex] does not converge (since it goes to infinity) but still have L=1.

    Is that right?
     
  8. Oct 8, 2009 #7

    Mark44

    Staff: Mentor

    Right.
     
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