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

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    $\begingroup$ Seen this? $\endgroup$
    – J. M. ain't a mathematician
    Sep 8, 2011 at 14:00
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    $\begingroup$ Because if rearrangements did not affect the infinite sum then the series would be either absolutely convergent or not convergent. $\endgroup$
    – Henry
    Sep 8, 2011 at 14:05
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    $\begingroup$ 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. :-) $\endgroup$
    – Srivatsan
    Sep 8, 2011 at 14:06
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    $\begingroup$ Related... $\endgroup$
    – J. M. ain't a mathematician
    Sep 8, 2011 at 14:13
  • $\begingroup$ @Henry Maybe I am being dense, but I see no way of showing/understanding your claim without going through the Riemann rearrangement theorem. $\endgroup$
    – Srivatsan
    Sep 8, 2011 at 14:24

3 Answers 3

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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.

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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.

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    $\begingroup$ Yours and martini I understand, HTH I think I managed to decode, but AB... $\endgroup$
    – Did
    Sep 11, 2011 at 16:24
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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.

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    $\begingroup$ Oops, I didn't see that the link was already given by J. M. $\endgroup$
    – George
    Sep 8, 2011 at 14:13

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