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So i am asked to rearranje the terms in this series:

$$ \sum_{i=1}^\infty \frac{(-1)^{n+1}}{n} = 1- \frac 12 +\frac13-\frac14+... $$

so that the sum of the series is equal to 0. I've seen the Riemann series theorem but i can't still can't find a rearrangement that satisfies the condition.

J. Barbosa
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From the Wikipedia article quoted in the comment, we get that the sum converges to $ln(2\sqrt{r})$ whenever $r$ is the ratio of positive to negative terms in the limit. Now, we want $ln(2\sqrt{r}) = 0$ so $2\sqrt{r} = 1$ so $r = \frac14$. This implies taking $1$ positive term for every $4$ negative terms will give you a sum that converges to $0$:

$1 - \frac 12 -\frac14 -\frac16 -\frac18 +\frac13 -\frac1{10} -\frac1{12} -\frac1{14} -\frac1{16} ... $

Colm Bhandal
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  • An interesting exercise now would be to check this answer. That is, find a simple proof that this series does in fact converge to $0$- without appealing to the Reimann series theorem. Maybe it's not possible, but it might be. – Colm Bhandal Jul 30 '15 at 14:48
  • Yes, it is: https://math.stackexchange.com/a/2277682/134791 – Jaume Oliver Lafont May 12 '17 at 10:05