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The harmonic series converges conditionally; therefore, the series can be rearranged in any way to get different sums, but how do we rearrange in such a way that it equates to 0. Is there a trick/ formula I could use to manipulate the series to equate it to 0 or to any number?

anonymous
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    The ALTERNATING series corresponding with the harmonic series can be arranged this way, not the harmonic series itself. – Peter Apr 04 '18 at 08:09
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    Roughly speaking add positive terms until you get a non-negative value, add negative terms until you get a non-positive value and so on. – Peter Apr 04 '18 at 08:12
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    I know this has come up here before. https://math.stackexchange.com/questions/911293/summing-various-rearrangements-of-1-frac12-frac13-frac14-cdots isn't exactly it, but helpful. – Gerry Myerson Apr 04 '18 at 08:45
  • See also https://math.stackexchange.com/questions/1623693/arranging-the-alternating-harmonic-series-to-sum-to-sqrt2 – Gerry Myerson Apr 04 '18 at 08:50

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I let do the task by Mathematica, and obtained a surprise:

enter image description here

We see here not a more or less random sequence of positive and negative terms, depending on arithmetic accidents, but a definite pattern: There are blocks of five subsequent terms of the following kind: $${1\over 2n-1}, -{1\over 8n-6}, \ -{1\over 8n-4}, \ -{1\over 8n-2}, \ -{1\over 8n} \qquad(n\geq1)\ .$$ I hope that someone will be able to prove that the sequence will go on forever like that.

Christian Blatter
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