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

[GeoMike]

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-

 

[Pere Callahan]

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.

 

[Aleph-0]

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-

Not exactly, it is known that the series $$\sum \frac{1}{n}$$ is divergent, while $$\sum \frac{(-1)^n}{n}$$ is convergent.
Both have terms which converges to zero, with the same "rate".
However, the last one is, of course, not absolutely convergent.

 

[ircdan]

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