3

In the case of conditional convergence of a series, why do rearrangements affect the value of the series?

J. M. ain't a mathematician
  • 75,051
Prigmin Smith
  • 407
  • 2
    Seen this? – J. M. ain't a mathematician Sep 08 '11 at 14:00
  • 4
    Because if rearrangements did not affect the infinite sum then the series would be either absolutely convergent or not convergent. – Henry Sep 08 '11 at 14:05
  • 3
    A nitpick. Note that a rearrangement changes every series, not just the conditionally convergence ones. You meant to ask: why do rearrangements affect the convergence (and/or sum) of the series? @J.M. I wanted to make that edit, but I thought it was significant enough to point it to the OP. Anyway. :-) – Srivatsan Sep 08 '11 at 14:06
  • 1
    Related... – J. M. ain't a mathematician Sep 08 '11 at 14:13
  • @Henry Maybe I am being dense, but I see no way of showing/understanding your claim without going through the Riemann rearrangement theorem. – Srivatsan Sep 08 '11 at 14:24
  • @Srivatsan: it was almost a joke response to a circular question - why is something what it is defined to be? – Henry Sep 08 '11 at 17:32
  • @Henry, now I am even more confused. There's nothing circular about the question. A conditionally convergent series is defined as a series that is convergent, but not absolutely convergent. The definition does not say anything about rearrangements. And the proof for either the question or your statement isn't that trivial either. In short, I am missing the joke. :-) – Srivatsan Sep 08 '11 at 17:37
  • @Srivatsan: Another definition might be a series where the value is conditional on the order of the terms. – Henry Sep 08 '11 at 19:10
  • @Henry I have never seen that definition before. I am talking about this. I couldn't verify your definition, but I found this. I don't know if my interpretation is the correct intepretation, but it definitely seems like a standard one. – Srivatsan Sep 08 '11 at 19:12

3 Answers3

11

The intuitive answer is that in a conditionally convergent series the positive terms sum to $+\infty$ while the negative terms sum to $-\infty$. We know $\infty-\infty$ is not well defined. If we sum up the positive terms faster than the negative ones we increase the value of the sum. The link J. M. gave is a good one.

Ross Millikan
  • 374,822
2

consider the series $\sum_{i=1}^\infty (-1)^{i+1}\cdot \frac1i$. It is (conditionally) convergent. We now construct a divergent rearrangement. Since $\sum_{i=1}^\infty \frac 1{2i+1}$ diverges, we can choose an increasing sequence $(N_k)$ with $N_0 = 0$ in $\mathbb N$ s.th. $\sum_{i=N_k+1}^{N_{k+1}} \frac 1{2i+1} \ge 1$ for all $k\in \mathbb N$. Now $$ \sum_{k=1}^\infty \left(\sum_{i=N_k+1} \frac 1{2i+1} - \frac 1{2i}\right) $$ is a divergent rearrangement of the given series. I hope, I understood your question correctly,

HTH, Yours, AB, martini.

martini
  • 84,101
1

This is called Riemann's rearrangement theorem. A better description than I could possibly type up here is given at wikipedia. It has detailed examples and a full proof.

George
  • 1,837