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Convergence and Divergence of a series

  • Thread starter mohabitar
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  • #1
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The series from n=1 to infinity log(n/(n+1)). This was on my quiz, which I got wrong. Here's what I did:
lim n-->infinity of log(n/(n+1))
so then that becomes: log(lim n-->infinity n/(n+1))
which becomes the log1, which is 0, so it converges.

Whats wrong with my steps?
 

Answers and Replies

  • #2
What you showed was that the sequence converges to 0, not the series. The sequence has to converge to 0 for the series to converge, but it does not guarantee that it does. A good example of this is the harmonic series
[tex]\sum_{k=1}^{\infty} \frac{1){k} [/tex]
The limit as k goes to infinity of 1/k is zero, but the series still diverges.
 
  • #3
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If you have a series [itex]\sum a_n[/itex], and lim an is not 0, or the limit doesn't exist, then you know that your series diverges.

If lim an = 0, then you really can't say much at all about your series. That's what all the tests (comparison, ratio, root, integral, limit comparison, etc.) are about.
 
  • #4
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I thought what I did was a test?? Ahhh this stuff is so confusing! So many times in my book it said if the lim of an=0, then it converges. I dont get whats going on.
 
  • #5
33,508
5,193
I thought what I did was a test?? Ahhh this stuff is so confusing! So many times in my book it said if the lim of an=0, then it converges. I dont get whats going on.
I suspect that you are confusing a sequence, {an}, with a series, [itex]\sum a_n[/itex]. If lim an = 0 (or any specific number), the sequence converges, but nothing can be said about the series [itex]\sum a_n[/itex].

The harmonic series that rakalakalili gave and this series [tex]\sum_{k=1}^{\infty} \frac{1}{k^2} [/tex]
are such that lim an = 0, but the harmonic series diverges and the other series converges.
 

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