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Every example of a conditionally convergent series I can think of is alternating. Can someone find a non-alternating conditionally convergent series? Thanks.

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  • $\begingroup$ Insert zeroes appropriately in any alternating series. $\endgroup$ – copper.hat Nov 7 '13 at 19:58
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Any convergent reordering of a conditionally convergent series will be conditionally convergent. A typical example is the reordering $$ 1,-\frac12,-\frac14,\frac13,-\frac16,-\frac18,\frac15,-\frac1{10},-\frac1{12},\frac17,-\frac1{14},\ldots $$ of the alternating harmonic series, with sum $\frac12\,\log2$.

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Here is a non trivial example: $$ \sum \frac{\sin n}{n} $$

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Trivial example:

$$1 + 0 - \frac{1}{2} + 0 + \frac{1}{3} + 0 - \frac{1}{4} + 0 + \dots$$

is a non-alternating version of the alternating harmonic series.


Not-as-trivial example: Consider a sequence of the form

$$\frac{1}{2}, \frac{1}{2}, -1,$$$$ \frac{1}{3}, \frac{1}{3} -\frac{2}{3},$$ $$\frac{1}{4}, \frac{1}{4}, -\frac{1}{2},$$

and so on. The series associated to this sequence converges (to $0$, in fact), but it's not absolutely convergent.

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  • $\begingroup$ Thanks. Another followup: is it true that all conditionally convergent series have an infinite number of negative terms? $\endgroup$ – r123454321 Nov 7 '13 at 20:03
  • $\begingroup$ @RyanYu Yes. For if all but finitely many terms are positive, just replace finitely many terms with their absolute values. Changing finitely many terms does not affect convergence. $\endgroup$ – user61527 Nov 7 '13 at 20:04
  • $\begingroup$ Yes. $\sum a_n=c$, $\sum |a_n|=\infty$. The difference is twice the sum of the negative terms, thus the sum of the negative terms is divergent, and thus there are infinitely many negative terms. $\endgroup$ – Tim Ratigan Nov 7 '13 at 20:06

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