How come it be $\frac{3}{2}A$ and not only $A$?

Question

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 ?

Answer 1

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 $\pm \infty$

Answer 2

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 $1 - \frac12 + \frac13 - \frac14 + - \ldots$.