Given the array, $$\begin{matrix} -1 & 0 & 0 & 0 & \ldots \\ 1\over2 & -1 & 0 & 0 & \ldots \\ 1\over4 & 1\over2 & -1 & 0 & \ldots \\ 1\over8 & 1\over4 & 1\over2 & -1 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \end{matrix}$$ Here $\sum_i\sum_ja_{ij} = -2$, while $\sum_j\sum_ia_{ij} = 0$. Why both sums are different?
why the two ways of adding elements of array yield different results?
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0Presumably the number at the bottom left was meant to be $1/8$? – 2012-03-11
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0@joriki, yes it is 1/8. – 2012-03-11
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2More might be said if you explain why you think the sums should be the same. – 2012-03-11
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1@joriki, I think I may have got it. It is because the series does not converge absolutely, hence the rearrangement might yield different values. – 2012-03-11
1 Answers
I assume you think that the two should be the same because you are adding all the same elements, simply ordering them differently. While it is true that rearranging an absolutely convergent series will give you the same result, no matter how you order the elements $a_{ij}$ in your double series (in order to make it a single series) you will have a conditionally convergent series, which can be rearranged to converge to any real number and even diverge. This fact is known as the Riemann rearrangement theorem, and is very cool.
Edit: As Robert Israel pointed out, you can't actually make this into a conditionally convergent single series as the limit of the terms can't go to $0$ (since there are infinitely many $-1$'s). So in this case the failure of rearrangement preserving the value of the series is even stronger.
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1Actually it will be a *divergent* series (note there are infinitely many $-1$'s). – 2012-03-11
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0@RobertIsrael Good point, editing. – 2012-03-11
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0For an example where you do get conditionally convergent series, try $$ \pmatrix{1 & 0 & 0 & 0 & \ldots \cr -\frac{1}{2\cdot 1} & \frac{1}{2} & 0 & 0 & \ldots\cr -\frac{1}{3\cdot2} & -\frac{1}{3\cdot 2} & \frac{1}{3} & 0 & \ldots\cr -\frac{1}{4\cdot 3} &-\frac{1}{4\cdot 3} &-\frac{1}{4\cdot 3} & \frac{1}{4} & \ldots\cr \ldots & \ldots & \ldots & \ldots & \ldots\cr}$$ – 2012-03-11