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Here is my attempt to understand this: Let $$ \sum_{n=1}^{\infty}\left(-1\right)^{\left(n+1\right)}a_n $$ be an alternating series

now the infinite sum is defined as the limit of the sequence of partial sums and so a partial sum of $m$ terms of the series where $m$ is an even number (I think it wouldn't make a difference if $m$ was odd since we will take the limit as $m$ approaches infinity anyway) is: $$ \sum_{n=1}^{m}\left(-1\right)^{\left(n+1\right)}a_n = \sum_{n=1}^{m/2}(a_{2n-1}) - \sum_{n=1}^{m/2}(a_{2n})$$

Now taking the limit as $m$ approaches infinity on both sides and denoting the limit on the left by $S$ , the first sum on the right side by $S_o$ and the last sum by $S_e$

$$S = \lim_{m\rightarrow\infty}(S_o-S_e)$$

the right term is the limit of a difference , If limits as $m$ approaches infinity of both $S_o$ and $S_e$ exist the limit can be separated which means that no matter how you change the order of summation of the original sum it can be rearranged such that the positive terms are together and the negative terms are together and since both limits exist their values are unique and their difference is the unique sum of the whole series.

this seems to make sense until I try to convince myself that if one (or both) of the limits of $S_o$ and $S_e$ as $m$ approaches infinity doesn't exist (since proving that one doesn't exist implies the same of the other) corresponds to proving the fact that changing the order of summation of the original series changes the value of $S$ and that it isn't unique anymore. I stop here with no idea how to proceed.

Thanks in advance

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  • $\begingroup$ The main idea is that the tail of your series will be small, even if you take absolut value of your terms. This will give you that the sequence of partial sums is a Cauchy sequence and thus the series will converge. $\endgroup$ – Severin Schraven Mar 30 at 11:40
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    $\begingroup$ The question isn't about whether the series converges or diverges , it's about the variance or invariance of the limit of the sum under changing the order of summation $\endgroup$ – Km356 Mar 30 at 18:23
  • $\begingroup$ The whole point is the convergence. Splitting in positive and negative elements and using what I wrote above implies that you can reorder without changing the value. $\endgroup$ – Severin Schraven Mar 31 at 16:10
  • $\begingroup$ You can also see it like this. First you show that reordering doesn't change the value of your series if all element are positive (or all negative). And then you consider $a^+_n=\max\{a_n, 0\}$ and $a^-_n=\min\{0, a_n\}$. Then we have $a_n=a^+_n + a^-_n$ and we can reduce to the case where all the elements have the same sign. $\endgroup$ – Severin Schraven Mar 31 at 16:16
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this seems to make sense until I try to convince myself that if one (or both) of the limits of $S_o$ and $S_e$ as $m$ approaches infinity doesn't exist (since proving that one doesn't exist implies the same of the other) corresponds to proving the fact that changing the order of summation of the original series changes the value of $S$ and that it isn't unique anymore.

A sum $A$ of an absolutely convergent series does not depend on the order of summation because a main contribution to $A$ is made by (finite) heads of the series, for which we can apply the commutativity of addition, whereas sums of tails of the sequence can be arbitrarytiry small, see a proof.

Conversely, if a series converges, but not absolutely then a sum of its chosen members can tend to infinity, but the fixed order of the series prevents infinite partial sum growth by its collapsing, for instance, in a series $\sum_{n=1}^{\infty}\left(-1\right)^{\left(n+1\right)}\tfrac 1n$. If we relax this order then we can force the rearranged series to converges to an arbitrary real number or to diverge, see Riemann series theorem.

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    $\begingroup$ To the proposer: For example if $a_n=1/n$ then $S_o(m)$ and $S_e(m)$ tend to $\infty$ as $m\to \infty$ but $S=\ln 2 \approx 0.693147.$ $\endgroup$ – DanielWainfleet Apr 8 at 3:11
  • $\begingroup$ I thought I'd share that if we rearrange the harmonic series as $ 1+1/3-1/2+1/5-1/4... $ then the sum $S = \frac{3}{2}\ln{2} $. This is an example of the statements of the Riemann series theorem. $\endgroup$ – Azhao17 Apr 8 at 18:42
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By definition, the sum of a series is the limit of a sequence of partial sums. The peculiarity of the conventionally convergent series is that it is possible to distinguish in it the terms for which this limit does not exist.

Let us consider the series $$s=\log\dfrac{1+1}1+\log\dfrac{2-1}2+\log\dfrac{3+1}3+\log\dfrac{4-1}4+\log\dfrac{5+1}5+\log\dfrac{6-1}6+\dots.$$

A sequence of partial sums $$\left\{\log\dfrac21, 0, \log\dfrac43, 0, \log\dfrac65,0,\dots\right\}$$ converges.
Partial sums of the positive subsequence are $$\left\{\log\left(\dfrac{2}{1}\right),\log\left(\dfrac{2\cdot4}{1\cdot3}\right),\log\left(\dfrac{2\cdot4\cdot6}{1\cdot3\cdot5}\right),\dots\right\}$$ and of the negative one are $$\left\{\log\left(\dfrac{1}{2}\right),\log\left(\dfrac{1\cdot3}{2\cdot4}\right),\log\left(\dfrac{1\cdot3\cdot5}{2\cdot4\cdot6}\right),\dots\right\},$$ both of them diverges.

Unproportial parts of the positive and negative terms allow to change the sum, doing transpositions of terms.

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