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$$\log2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$ $$\frac{\log2}{2}=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots$$ Adding these two convergent series gives $$\log2 + \frac{\log2}{2}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\cdots+2\left(-\frac{1}{4}-\frac{1}{8}-\frac{1}{12}-\frac{1}{16}\cdots\right)=\log{2}$$

What did I do wrong?

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Such series, which are not absolutely convergent, when rearranged give different values with each rearrangement. – John Dec 20 '13 at 7:06
Be careful about rearrangements – mathlove Dec 20 '13 at 7:07
Neither of the two series in the middle of your last display equation are even convergent---let alone non-absolutely convergent. What you have is of the form $\infty -\infty$. – John Bentin Dec 20 '13 at 9:57
@OJB See the Wikipedia article. – Alyosha Mar 18 '14 at 21:07
up vote 14 down vote accepted

The series involved are not absolutely convergent, so they cannot be rearranged with the same result; that is, there isn't an infinite associative property for conditionally convergent series.

In fact, it's generally true that a series is absolutely convergent if and only if all its rearrangments converge to the same limit. It's also true that, given a real number and a conditionally convergent series, there is a rearrangement of the series convering the to the number.

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Well that's annoying, I wasn't aware of this. I'll look into it some more, thanks. – bwv869 Dec 20 '13 at 7:12
@Oliver It's fun! It certainly does make it necessary to use caution, though. – user61527 Dec 20 '13 at 7:13

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