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I just learnt that one of the axioms of a summation method for a divergent series is linearity: $$S[\sum_{n=0}^{\infty}(\alpha a_{n} + \beta b_{n})] = \alpha S[\sum_{n=0}^{\infty} a_{n}] + \beta S[\sum_{n=0}^{\infty} b_{n}]$$

However this stroke me as changing the order of the series (using associativity and commutivity), which the teacher was saying doesn't apply to divergent series! It seems like you are doing this: $$(\alpha a_{0}+\beta b_{0})+(\alpha a_{1}+\beta b_{1})+... \longrightarrow (\alpha a_{0}+\alpha a_{1}+...)+(+\beta b_{0}+\beta b_{1}+...)$$

In fact the series $1-1+1-1+...$ which by this general method sums to $\frac{1}{2}$ can be manipulated to sum to $0$ with this property it seems: $$S[\sum_{n=0}^{\infty}(1 + (-1) 1)] = S[\sum_{n=0}^{\infty} 1] + (-1) S[\sum_{n=0}^{\infty} 1]=0$$

What is the problem here?

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    $\begingroup$ Your example series is already zero and not 1-1+1-1.. I think you missed a power. Another tip is to make the upper bound not infinity yet but a variable. Then split it correctly. $\endgroup$ – Gerben May 21 '16 at 8:33
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The problem here is that you have to distinguish between the series $$ \sum_{n=0}^\infty a_k, \quad a_k = (1 + (-1)) = 0 $$ and the series $$ \sum_{n=0}^\infty b_k, \quad b_k=(-1)^k \text{.} $$

If you extend series summantion to non-convergent series, summantion methods simply are (pretty much arbitrary) maps $S$ from the set of series, i.e. $\mathbb{R}^\mathbb{N}$ to the set of real numbers $\mathbb{R}$. You usually put two requirements on those maps, namely that

  1. They are linear, in the sense that $S\left(\alpha (x_n)_{n\in\mathbb{N}} + \beta (x_n)_{n\in\mathbb{N}}\right) = \alpha S\left((x_n)_{n\in\mathbb{N}}\right) + \beta S\left((x_n)_{n\in\mathbb{N}}\right)$

  2. They map convergent series to their usual limit, i.e. for example $S\left(\left(2^{-n}\right)_{n\in\mathbb{N}}\right) = 2$.

Those requirements, however, don't imply that you may re-group summands. But you have to do that to map $(a_k)_{k\in\mathbb{N}}$ to $(b_k)_{k\in\mathbb{N}}$.

If I recall correctly, you can find a divergent (in the usual sense) series $(a_k)$ such that for every $l \in \mathbb{N}$ there's a re-ordering $(b_k)$ of $(a_k)$ such that $\sum_{k=1}^\infty b_k$ converges in the usual sense to $l$. Thus, extended summation methods, if they want to fullfill (2) above, have to be very sensitive to reorderings.

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  • $\begingroup$ True!. What do you say about how this property (forgetting about the multiplicative factors) seem to be like a certain reordering as I said in the question? $\endgroup$ – guillefix Apr 27 '14 at 15:21
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    $\begingroup$ @guillefix Well, you can view linearity as a rather special kind of reordering. But what does that buy you? $\endgroup$ – fgp Apr 27 '14 at 19:55
  • $\begingroup$ Well it was just to see what's the motivation behind linearity. The multiplication factor I see as just extending the distributive property to infinite sums, but the rest seems like accepting a certain reordering while you still don't accept others. My question is then is there a special reason why that reordering should be part of the definition of a 'sum'? $\endgroup$ – guillefix Apr 27 '14 at 21:21
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    $\begingroup$ @guillefix Hm, as you say you certainly want $\sum \alpha a_k = \alpha \sum a_k$. I don't see an immediate reason why $\sum a_k + b_k = \sum a_k + \sum b_k$ is absolutely necessary - but note that requiring this makes infinite summation methods linear functionals on the vector space of infinite sequences, which at the very least will have technical advantages, I believe. $\endgroup$ – fgp Apr 27 '14 at 21:58
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    $\begingroup$ It's worth noting that reordering changes the value of conditionally convergent series too. $\endgroup$ – user76284 Jul 17 at 21:45
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Your confusing "series" and a series of numbers. 1+1+1.. is not the same as 1+0+1+0+1. So please write the function your adressing and not the results you wish to sum over. You are regulating the function! and not the numbers. This is the rule that differ from finite summations.

If you have the series: 1-1+1-1... the best is to write what you actualy mean: $$\sum_{n=1}^{\infty} (-1)^{n+1}= 1/2$$

If you split them in 1+0+1+0 and 0-1+0-1 etc. By linearity you will get the same result.

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}+1}{2}=0$$ $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}-1} {2}=1/2 $$

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