Change of order of summation in infinite series

I am having some difficulty to understand how to interpret the indices when talking about change of order of summation in infinite series. For example, I have, for finite set of indices $F$ the following equality: $\sum_{k\in F} a_k = \sum_{F} a_k$. I am not sure I understand what the difference in both notations is. I see both of them as running on all the terms in $F$ and adding all of them together. So why do we need two notations? Or is it just a definition? (what does it mean "the sum above is well-defined"?).

I also have a theorem that says: Let $I=\cup_{j=1}^{\infty}I_j$ where $I_j\cap I_k=\emptyset$ for $j\neq k$. If $a_i \geq 0$ and $i\in I$ then $\sum_{I} a_i = \sum_{j=1}^{\infty}\left( \sum_{I_j} a_i \right )$.

I am not sure how to interpret the indices and how to read it. Eventually it should be that for positive series, the order of summation doesn't matter, but I can't see it. Can't do the do step from the theorem to the interpretation.

I will be grateful if someone can help me understand how to read it.

• If you want more details on how to read it I can provide them... Anyway the theorem is valid only if $a_n$ is absolutely convergent. – N74 Sep 5 '16 at 19:58

For finite $F$ the only difference between $\sum_{k\in F}a_k$ and $\sum_Fa_k$ is notational: the latter is an abbreviation for the former.

The theorem does indeed say that a non-negative series can be rearranged and broken up arbitrarily without any effect on its value. This is perhaps easiest to see in the special case in which $I=\Bbb Z^+$, say, so that

$$\sum_Ia_k=\sum_{k\in\Bbb Z^+}a_k=\sum_{k=1}^\infty a_k\;.$$

Call this sum $S$. Now we partition the positive integers into sets $I_j$ for $j\in\Bbb Z^+$: these sets are pairwise disjoint, and their union is all of $\Bbb Z^+$. The theorem says that if for each $j\in\Bbb Z^+$ we set

$$S_j=\sum_{I_j}a_k=\sum_{k\in I_j}a_k\;,$$

then

$$S=\sum_{j=1}^\infty S_j\;.$$

We might, for instance, let $I_1=\{2\}$, $I_2=\{1\}$, $I_3=\{4\}$, $I_4=\{3\}$, and so on, so that

$$I_j=\begin{cases} \{a_{j+1}\},&\text{if }j\text{ is odd}\\ \{a_{j-1}\},&\text{if }j\text{ is even}\;; \end{cases}$$

then

$$S_j=\begin{cases} a_{j+1},&\text{if }j\text{ is odd}\\ a_{j-1},&\text{if }j\text{ is even}\;, \end{cases}$$

and

$$\sum_{j=1}^\infty S_j=a_2+a_1+a_4+a_3+a_6+a_5+\ldots$$

is just the sum of the rearrangement in which we interchange $a_{2n-1}$ and $a_{2n}$ for each $n\in\Bbb Z^+$. Any rearrangement of the original series can be obtained in this way, and the theorem says that they all produce the same sum.

But it actually says even more than that, since the sets $I_j$ can contain more than one index. For instance, we can let

$$I_1=\{2n-1:n\in\Bbb Z^+\}=\{1,3,5,7,\ldots\}$$

and

$$I_2=\{2n:n\in\Bbb Z^+\}=\{2,4,6,8,\ldots,\}\;,$$

setting $I_j=\varnothing$ if $j>2$. Then

\begin{align*} S_1&=\sum_{n=1}^\infty a_{2n-1}=a_1+a_3+a_5+\ldots\;,\\ S_2&=\sum_{n=1}^\infty a_{2n}=a_2+a_4+a_6+\ldots\;, \end{align*}

and $S_j=0$ for $j>2$. Clearly we can ignore the $0$ terms, so the theorem says in this case that $S=S_1+S_2$, i.e., that

$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty a_{2n-1}+\sum_{n=1}^\infty a_{2n}\;:$$

we can sum the odd-numbered and the even-numbered terms separately, and the sum of those two subseries totals will be the same as the sum of the original series. Thus, the theorem covers not only rearrangements of the individual terms, but also breaking up the series into disjoint subseries and summing those subseries individually.

I can’t say exactly what is meant by the sum above is well-defined without having the entire relevant context. If it refers to $\sum_{k\in F}a_k$ for a finite $F$, for instance, it simply means that since this is a finite sum, we already know that the order in which it’s evaluated makes no difference, so we can safely specify the set of indices as a ‘lump’ — the set $F$ — instead of specifying the order in which the terms are to be added.

• Wow, thank you for the explanation. The sentence "the sum above is well-defined" was referred to the finite sum $\sum_{k\in F} a_k$. When you say $\sum_{k\in F} a_k$, it doesn't mean that $k$ should first be 1, then 2 and so on, but you can run through F in arbitrary order? – Mr. Tea Sep 6 '16 at 16:47
• because you wrote $\sum_{k\in F} a_k =\sum_{ F} a_k=\sum_{k=1}^{\infty} a_k$, which if I understand correctly what the theorem says. – Mr. Tea Sep 6 '16 at 16:52
• @Mr.Tea: No, $\sum_{k\in F}a_k$ just means the sum of the finite set of numbers $a_k$ with $k\in F$; no specific order is implied. The sum is still well-defined even without an implied order, because we can prove that the order doesn’t make a difference for finite sums. The whole discussion that you’re reading arises because it turns out that it can make a difference for infinite series — specifically, when they are only conditionally convergent. The theorem here is showing that it nevertheless doesn’t make a difference when all of the terms are non-negative. – Brian M. Scott Sep 6 '16 at 16:55
• I am not sure why you can say that $\sum_{k=1}^{\infty}a_k=\sum_{k \in \mathbb{Z}^+}a_k =S$. Doesn't it imply from the theorem? Sorry for all the questions... – Mr. Tea Sep 6 '16 at 17:44
• @Mr.Tea: $\sum_{k=1}^\infty a_k=\sum_{k\in\Bbb Z^+}a_k$ by definition: in the case of an infinite series we do assume the usual order of terms when we write something like $\sum_{k\in\Bbb Z^+}a_k$. – Brian M. Scott Sep 6 '16 at 17:47