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Divergent series and the limit of the nth term as n approaches infinity

  1. Apr 10, 2008 #1
    I'm looking for help with my conceptual understanding of part of the following:

    1) If a series is convergent it's nth term approaches 0 as n approaches infinity
    This makes perfect sense to me.

    2) If the nth term of a series does not approach 0 as n approaches infinity, the series is divergent
    Again, makes perfect sense.

    3) A divergent series can have an nth term that approaches 0 as n approaches infinity. Thus #1 cannot be used as a test FOR convergence.
    Here's where I'm thrown a little. I can follow the proofs in my textbook fine, and I think I see what they all suggest.
    Essentially: The RATE at which the terms of a series approaches zero (assuming they do at all) is what really determines convergence/divergence -- am I understanding this right?

    Thanks,
    -GM-
     
  2. jcsd
  3. Apr 10, 2008 #2
    Yes you're right. The sequence has to tend to zero fast enough for the sum to converge.

    For example the sum of 1/n does not converge altough 1/n goes to zero, but too slowly, the same for 1/log(n). In fact as you know the sum of 1/n^s converges for all s>1 and diverges for s<=1 so this gives you an idea of how fast the sequence should go to zero.
     
  4. Apr 10, 2008 #3
    Not exactly, it is known that the series [tex]\sum \frac{1}{n}[/tex] is divergent, while [tex]\sum \frac{(-1)^n}{n}[/tex] is convergent.
    Both have terms which converges to zero, with the same "rate".
    However, the last one is, of course, not absolutely convergent.
     
  5. Apr 10, 2008 #4
    another example, sum((-1)^n/sqrt(n)) also converges, but again, not absolutely, actually we can put anything in the denominator with an n, even sum((-1)^n/n^(1/1000)) converges
     
    Last edited: Apr 10, 2008
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