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Can a series have a limit and be divergent? I'm confused about the difference between divergence and a series not having a limit. A ck-12 calculus book stated they are different concepts.

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The limit could be $\infty$ for example, in which case we still say it's divergent. – Najib Idrissi Jun 18 '13 at 19:04
If you are reading this from a calculus book, make sure you are talking about the same objects having a limit. For example, a series $\sum_{n = 1}^\infty a_n$ converges if and only if the sequence of partial sums $s_k = \sum_{n = 1}^k a_n$ has a limit as $k$ approaches $\infty$. This is not the same as determining if the sequence $a_n$ has a limit as $n \to \infty$. – Ben Passer Jun 18 '13 at 21:09

If a series is convergent then it has bounded partial sums. Equivalently, if a series fails to have bounded partial sums then it is divergent.

The converse is false. If a series has bounded partial sums then it need not be convergent. A counter-example is $\sum_{n \ge 1}(-1)^n$ where the partial sums are $-1,0,-1,0,-1,\ldots$.

The partial sums obtain a finite limit if and only if the series is convergent.

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Boundedness of partial sums is a sufficient condition if the terms are nonnegative, though. – Math1000 May 27 '15 at 13:57

In every calculus and real analysis book I've read, the two are the same thing.For example, in Stewart's calculus:

Either a series converges, or it diverges. Converge means it has a limit. Diverge means it doesn't.

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