If $\sum(-1)^{n+1} \frac1n$ can be made to sum to any number then why is it equal to $\ln 2$?

Question

I have two questions, one is in the title:

If $\sum(-1)^{n+1} \frac1n$ can be made to sum to any number then why it is equal to $\ln 2$? Certainly it occurs since the divergence of each of the two subseries (negatives and positives) guarantees that we will always be able to equate that to any desired number but why we choose $\ln 2$ instead?

For example, $\ln 2 = 1- \frac12 + \frac13 - \frac14 + \dots = 1- \frac12 - \frac14 + + \frac13 - \frac16 - \frac18 + \frac15 - \dots = \frac12 \ln 2 $

And second, which necessary and sufficient conditions a series must have to converge to one specific number if it converges to a number at all? And what proof is that for those necessary and sufficient conditions?

Answer 1

The result you are using is called the Riemann Rearrangement Theorem. It says that if a series is conditionally convergent, then for any real number there is a rearrangement of the series such that the new series converges to that number. There are two points to note here:

1. A series is not just a sequence of numbers that is summed up. It is a sequence of numbers that is summed up in a specific order. So, the rearrangements are all distinct series. They are not the same series, so it is incorrect to say that $\sum (-1)^n / n$ can be summed up to any real value. Each rearrangement of this series is a different series that converges to a unique real number, if it converges at all.
2. It is crucial that the series you consider be conditionally convergent. An absolutely convergent series is convergent; moreover, any rearrangement of an absolutely convergent series is absolutely convergent, and it converges to the same value as the original series. This gives you the necessary and sufficient conditions you wanted.

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You can find a proof of the Riemann Rearrangement Theorem in any standard textbook on real analysis, say Walter Rudin's Principles of Mathematical Analysis.

It is a straightforward exercise to show using an $\epsilon$-$\delta$ argument that an absolutely convergent series is convergent. It may be slightly more difficult to show that any rearrangement of an absolutely convergent series converges to the same value, but this site has covered that scenario as well.

Answer 2

It can be made to sum to any number by a suitable rearrangement of terms. In the order $1-\frac12+\frac13-\frac14+\cdots$, it has the unique value $\ln 2$; that is to say, by evaluating the sum of the first $n$ terms in this order, we can get as close to $\ln 2$ as we like by making $n$ large enough.