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Infinite series: nth-term test

  1. Oct 22, 2004 #1
    I'm reading How to Ace the Rest of Calculus and on page 31 there's a test for divergence that says if the limit (as n goes to infinity) of a_n is NOT equal to zero, then the infinite series [tex]\sum a_n[/tex] (that goes from n = 1 to infinity) diverges.

    3 pages before that, on 28, there's an example on a series that converges to 1... The sum is represented by [tex]S_n = (2^n - 1)/2^n[/tex]

    I'm confused here: since taking the limit of that [tex]S_n[/tex] equals to 1... shouldn't the series diverge? Come to think of it I know I'm missing something here because as I understand it according to that test ANY series that has a limit to any number but 0 diverges which makes no sense. I know that test does nor work in reverse btw (limit going to 0 doesn't mean it converges) but this still sounds like all series convergence must go to 0... which I know is wrong. :confused:
     
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  3. Oct 22, 2004 #2

    arildno

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    You are confusing [tex]a_{n}[/tex] with [tex]S_{n}[/tex]
    [tex]a_{n}[/tex] is just some sequence of numbers, while [tex]S_{n}[/tex] is the sum of the first n [tex]a_{j} (j=1,2...n)[/tex]
    Was that clear enough?
     
  4. Oct 22, 2004 #3

    shmoe

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    With the divergence test you look at the limit of the terms in your series, the [tex]a_n[/tex]'s. This is different from the limit of the partial sums, the [tex]S_n[/tex]'s. Note [tex]S_n=\sum_{i=1}^{n}a_i[/tex] and we define [tex]\sum_{i=1}^{\infty}a_i=\lim_{n\rightarrow\infty}S_n[/tex]

    The example you gave, the partial sums [tex]S_n = (2^n - 1)/2^n[/tex], came from the terms [tex]a_n=1/2^n[/tex]. Here [tex]\lim_{n\rightarrow\infty}a_n=0[/tex] so the divergence test is inconclusive. The partial sums converge to 1, so we say the series converges.
     
  5. Oct 22, 2004 #4
    haha that was simple. :redface:

    Thanks. :smile:
     
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