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I am to prove that if $$1-1/2+1/3-1/4+1/5-1/6+...=s$$ that $$1-1/2+1/3+1/5-1/4+1/7+1/9-1/6+...=3s/2$$

My approach:

$s/2=1/2-1/4+1/6-1/8+1/10-1/12...$ so that

$s+s/2=3s/2=1-(1/2-1/2)+1/3-(1/4+1/4)+1/5-(1/6-1/6)...=$ $1+1/3-1/2+1/5...$

This sort of leads to what I want to approach, but the terms are not quite in the right order. I assume I may not change the order of the last set of terms for it to match what I'm looking for. Any other suggestions?

Marc
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  • I don't really understand what's being answered in that question/post... – Marc Dec 21 '17 at 21:03
  • i think this has to do with convergent, absolute convergent and conditionally convergent series: https://math.stackexchange.com/questions/640493/why-can-infinite-series-be-summed-different-ways-to-get-different-results and https://math.stackexchange.com/questions/357224/absolutely- and convergent-series-and-conditionally-convergent-series-rearrangement and https://math.stackexchange.com/questions/911293/summing-various-rearrangements-of-1-frac12-frac13-frac14-cdots?rq=1 – cgiovanardi Dec 22 '17 at 19:37

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