Convergence and Divergence of a series (1 Viewer)

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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?
 
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.
 
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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.
 
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.
 
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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.
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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