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$\sum u_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ conditionally converges to $\log 2$; and, when deranged to $\sum u'_n = 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\dots,$ converges to $\frac{1}{2}\log 2$. Why does a deranged conditionally convergent series have a different limit?

First Course in Math Anal, p.93, by Burkill, explains that the sums $s_n, t_n$ of the first n terms of the two series are different functions of $n$. This reasoning is incomplete, because different functions of $n$ can converge to the same limit; for example, let $r_n$ be the first $2n$ terms of $\sum u_n$, then $s_n$ and $r_n$ are different functions; besides, by differently defining terms, i.e. clustering pieces of a series, e.g.

\begin{align*} 1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}+\dots = \left(1-\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots \\ = \left(1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots,\\ 1-1+1-1+\dots = (1-1)+(1-1)+\dots \\ =(1-1+1-1)+\dots,\\ 1+1+1+1+\dots = (1+1)+(1+1)+\dots \\ = (1+1+1+1)+\dots, \end{align*}

one gets the sum of first n terms as different functions, but the series's limit remains.

One could view a changed limit from the perspective of the changed number of terms, Denote 'countably many', the size of natural number, as $\lvert\mathbb{N}\rvert$, then $\lvert\mathbb{N}\rvert = \lvert k\mathbb{N}\rvert = \lvert\mathbb{N}\times\mathbb{N}\rvert$. Then two series with different numbers (e.g. $\lvert \mathbb{N}\rvert$, $\lvert k\mathbb{N}\rvert$) of terms– and so having different limits– paradoxically have same number of terms–and so are possibly formed by the same terms. The paradox could be the fundamental reason.

Are there other explanations? Is the use here of 'countably many' is proper?

Arctic Char
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Charlie Chang
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  • The problem is that the $\log2$ expansion is not absolutely convergent i.e it keeps jumping around the point of convergence. You can rearrange the terms in the alternating harmonic series to get any real value. – Sam Jul 21 '20 at 05:32
  • If the series converges absolutely, then any rearrangement has the same value. I am not sure if your attempt is successfully distinguishing the case of conditionally convergent series from that of absolutely convergent series. – Sangchul Lee Jul 21 '20 at 06:21

2 Answers2

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This phenomenon occurs with conditionally convergent series, i.e. a convergent series where the absolute value of the terms will sum to $\infty$. In the case of $\sum (-1)^n / n$, the absolute values of the terms form the series $\sum 1/n$, which is famously divergent.

The Riemann rearrangement theorem tells us that, given any limit $L \in [-\infty, \infty]$, any conditionally convergent series can be rearranged to converge to $L$. It can also be rearranged to diverge.

How does it work? Conditionally convergent series can be seen as a "sum" of two divergent series, one positive series which diverges to $\infty$, and a negative series which diverges to $-\infty$. We pull terms from each series in order to manipulate the partial sums to be close to $L$.

If, say, $L$ is positive, take a bunch of positive series terms until the partial sum is now over $L$ (this has to happen in finitely many steps, since the positive terms sum to $\infty$). Now, start taking negative terms until the sum is less than $L$, then take positive terms, etc.

The partial sums will oscillate about $L$, and the swings in the oscillations will dampen as time goes on, since the terms of the series must tend to $0$. Because you always wait for the partial sums to swing strictly above $L$, then strictly below $L$, and repeat, you will eventually use every term of the positive series and the negative series, giving us a permutation of the original series.

I see it intuitively like a tug-of-war. The positive series pulls the partial sums towards $\infty$, and the negative series pulls the sums to $-\infty$. They are both infinitely powerful, and keep each other in check, relatively speaking. But, if the referee forces one side to stop pulling temporarily, the other side will gain a lot of ground. Once the side starts pulling again, they will be relatively well-matched. If the referee starts micromanaging everything, they can make the knot in the middle of the rope end up wherever he wants!

user810049
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If the sequence is conditionally convergent then the sum of the positive terms is infinite and the sum of the negative terms is infinite, but the terms themselves shrink to zero. If you choose any real number $a$, you can add positive terms in order until the sum is greater than $a$, then negative terms until the sum is less than $a$, then positive terms again until the sum is again greater than $a$, and so on. For any $a$ this method create a sequence conditionally convergent to $a$.

Peter
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  • That makes sense in that, I guess, it's like two progressions competing with each other, and a key to the answer is that at any given moment, one always moves faster than the other; it's indeed the fact that we can stop or 'freeze' the progressions (i.e. 'CUT' the series) --NOT the CLUSTERING of pieces of a series into items--and that whenever we do so we find to certain extent (it varies with different arrangements) one progression surpasses the other that makes the difference ('accumulates' the limits). In other words, if we can't 'cut' the series then the series' sums won't be different. – Charlie Chang Jul 21 '20 at 06:26
  • But I'm still thinking of why the cutting matters, given after all pieces of the two series are the same. – Charlie Chang Jul 21 '20 at 06:27
  • It is because the positive and negative sums are both infinite. You could also express the result as "infinity minus infinity can be anything". – Peter Jul 21 '20 at 06:30