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this is the image

OK I admit I was too lazy to type this question so I took a screenshot , I got it from the site @brilliant.org where it asked in terms of $A$ what would be the 2nd summation equation ? The explained solution was there no doubt there but I guess it was too good to enter my fat-head, so I decided to put this up here.

Now coming to the problem above if we expand $A$ we get $(1 - 1/2) + (1/3 -1/4) + ......$ and when we expand the 2nd summation equation we get the same thing repackaged that's $(1 + 1/3 - 1/2) + (1/5 + 1/7 - 1/4) + .....$ So aren't they necessarily same ? since they are summed up to infinity, and yet the 2nd summation series above is $\frac{3}{2}A$ and not $A$ according to the answer, so where do I err ?

Arnav Das
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    When you add infinite series the order of how you add the terms matters, surprisingly. Even if the numbers in each sequence is the same. – Nicolas Bourbaki Jan 09 '16 at 19:08
  • In fact, if the series, like the ones here, converge conditionally but not absolutely, we can produce any real limit we like by rearranging the terms appropriately. – Travis Willse Jan 09 '16 at 19:20

2 Answers2

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They are conditionally convergent series. The definition is a series that converges but if you take the absolute value of all the terms the series does not converge. If you take the absolute value of all the terms in the first, you get the harmonic series which diverges like $\log N$. When summing a conditionally convergent series the result is not the same when the order of the terms is rearranged. In fact, with a proper rearrangement you can get any result you want, including $\pm \infty$

Ross Millikan
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The series $1 - \frac12 + \frac13 - \frac14 + - \ldots$ is not absolutely convergent. It is a theorem that if the series converges, but not absolutely, we can rearrange the sequence of the terms to make the series converge to whatever value we like.

The second series in the question is just a particular rearrangement of the series $1 - \frac12 + \frac13 - \frac14 + - \ldots$.

David K
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