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?

  • $\begingroup$ why can it be made to sum to any number? Btw the $ln2$ part comes from the Taylor expansion of the logarithm. $\endgroup$ – Μάρκος Καραμέρης Aug 26 '18 at 19:53
  • 1
    $\begingroup$ If you rearrange the terms you may take any number you want. $\endgroup$ – dmtri Aug 26 '18 at 19:53
  • 3
    $\begingroup$ See Riemman theorem $\endgroup$ – dmtri Aug 26 '18 at 19:54
  • 1
    $\begingroup$ A sum $\sum_{n=1}^\infty a_n$ is supposed to be summed in a particular order: start with $a_1$ then add $a_2$ then add $a_3$ etc. This converges to $\ln(2)$ here. If you change the order you can change the sum if the series if conditionally convergent. $\endgroup$ – Winther Aug 26 '18 at 19:55

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.

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.

  • 1
    $\begingroup$ This is a perfect answer (normally I don't get nice answers) however I need to learn the Riemann Rearrangement Theorem from the mentioned book soon.. Thanks a lot :) $\endgroup$ – user231343 Aug 26 '18 at 20:07
  • 1
    $\begingroup$ @Edi No problem, glad to be of help. :) $\endgroup$ – Brahadeesh Aug 26 '18 at 20:09
  • 1
    $\begingroup$ The idea behind the proof is pretty simple. If you want to get your series to sum to a given $a$, take the sum of the positive terms in order until it is the sum is just above $a$; then add on the negative terms in order until the sum is just below $a$; and continue. $\endgroup$ – Jair Taylor Aug 27 '18 at 1:41
  • $\begingroup$ +1 and this becomes a "nice answer" officially. $\endgroup$ – Paramanand Singh Aug 27 '18 at 5:56
  • $\begingroup$ @ParamanandSingh high praise, thanks a lot! :) $\endgroup$ – Brahadeesh Aug 27 '18 at 6:07

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.

  • $\begingroup$ Yes I know from where ln(2) comes but I asked why if there are any number so why ln(2)? .. which @ Winther 's comment answers that partially. $\endgroup$ – user231343 Aug 26 '18 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy