# Interchange finite and infinite sum

Under which condition is it valid to interchange a finite and an infinite sum?

We have used

$$\sum_{x \in I} \sum_{y=0}^{\infty} f_{x,y}= \sum_{y=0}^{\infty} \sum_{x \in I} f_{x,y}$$

for a finite set $I$. Is this always valid or only under certain circumstances? Under which conditions would it be valid for an infinite (countable) set $I$?

Thank you very much.

• Check out Fubini-Tonelli theorem. Can look even here math.stackexchange.com/questions/466757/… – A.Γ. Jul 11 '15 at 13:59
• So whenever $f_{x,y} \geq 0$ for all $x,y$, we can interchange them, did I get that right? And this also holds whenever $I$ is infinite? But I cannot interchange them, if they are negative, even if one sum is finite? If both sums are finite, then I can interchange them, no matter whether they are non-negative or negative. Is this correct? – user136457 Jul 11 '15 at 14:05
• If $f_{x,y}\ge 0$ (or $\le 0$) then you can interchange them always (Tonelli result), however, you can get $+\infty$ (resp. $-\infty$), but "LHS is infinite iff RHS is infinite". If $f_{x,y}$ may have opposite signs then you need other conditions on $f_{x,y}$, for example, the absolute convergence of the double sum (Fubini result). Here you never get infinity. – A.Γ. Jul 11 '15 at 14:43
• The fact that one sum is finite does not help you much compared to an infinite sum. Interchanging with a finite sum correspond to a different shuffling of the series terms. It is known that terms of a conditionally convergent series (i.e. convergent, but not absolutely) can always be shuffled to produce any number. If both sums are finite there is no trouble at all to switch them. – A.Γ. Jul 11 '15 at 14:55
• Somehow I think this contradicts answer 1, in 1, doesn't it? Because there is written that one finite sum is always enough. – user136457 Jul 11 '15 at 15:29

Well, your second question is just answered by a particular case of the Fubini's Theorem. Let me explain it:

1 - You may always interchange a finite sum with an infinite one. This is just a consequence of linearity of the integral. When you have a countably infinite sum, you may take it as an integral, and one of the first properties of an integral seen on a measure theory course will be that, for $f_1,...,f_n$ 'fine' functions, then

$$\sum _k \int f_k = \int \sum_k f_k$$.

2 - Your second question is a more interesting, and it is completely answered by Fubini's theorem. There is a much simpler version for sequences, though, and it states that:

If a doubly-indexed infinite sequence $\{x_{n,m}\}$ is absolutely summable in some way, i.e., if, for exemple,

$$\sum _m \sum_n |x_{m,n}| < + \infty$$

Then you may change the order of summation and obtain the same result. This also holds for only positive sequences, and the result is the same.

As I already mentioned, this is part of a general Theorem called Fubini's Theorem, and I leave it up to you to check it, if interested.

Even more can be said in this case: in this case, you may even sum the sequence in a 'random' way, and you are still going to get the same result, where by "random way" we mean that, for every permutation $\sigma : \mathbb{N} \times \mathbb{N} \to \mathbb{N} \times \mathbb{N}$, then the first sum is also equal

$$\sum_{i,j} x_{\sigma(i,j)}$$

• Thank you very much. Do you mind pointing out the main difference between Fubini-Tonelli, Monotone Convergence and Dominated Convergence: When in my lecture notes Dominated Convergence is written, my professor always says it is due to Fubini-Tonelli. Is one of those stronger than the others? – user136457 Jul 11 '15 at 14:16
• You're welcome, always nice to help! Well, I work with mathematical analysis, and as my short experience has shown me, Fubini's theorem may be used to help proving that the Dominated Convergence Theorem holds in specific cases. But the latter is much easier to prove than the former. For example, in my Measure Theory course, I think we saw the DCT on the second or at most third class, while we were only able to prove Fubini's theorem after a half of the course. – João Ramos Jul 11 '15 at 14:21
• Thank you very much! This helped a lot. – user136457 Jul 11 '15 at 14:24

João's answer only addresses the case where each of the sequences $\{f_{x,y}\}_{y=0}^\infty$ is integrable, that is, absolutely convergent. If these series are not absolutely convergent, you cannot (and should not) think of them as Lebesgue integrals: the limit depends on the order of summation! However, a finite sum may also be interchanged with an infinite one if the series are only conditionally convergent.

Proposition 1. Let $I$ be non-empty and finite, and suppose that for each $x\in I$ the series $\sum_{y=0}^\infty f_{x,y}$ converges. Then the series $\sum_{y=0}^\infty \sum_{x\in I} f_{x,y}$ also converges, and we have $$\sum_{x\in I} \sum_{y=0}^\infty f_{x,y} = \sum_{y=0}^\infty \sum_{x\in I} f_{x,y}.$$ Proof. We have \begin{align} \sum_{x\in I} \sum_{y=0}^\infty f_{x,y} &= \sum_{x\in I} \left(\lim_{n\to\infty} \sum_{y=0}^n f_{x,y}\right)\tag*{(by definition)}\\[1ex] &= \lim_{n\to\infty} \left(\sum_{x\in I} \sum_{y=0}^n f_{x,y}\right)\tag*{(since addition is continuous)}\\[1ex] &= \lim_{n\to\infty} \left(\sum_{y=0}^n \sum_{x\in I} f_{x,y}\right)\tag*{(interchanging finite sums)}\\[1ex] &= \sum_{n=0}^\infty \sum_{x\in I} f_{x,y}.\tag*{$\Box$} \end{align}

The proof shows that the result holds in any abelian topological group (or even semigroup), the minimum structure needed to talk about infinite series. (For instance, this means that you can use it in topological vector spaces without having to resort to vector-valued integration.)

We cannot omit the assumption that for each $x\in I$ the series $\sum_{y=0}^\infty f_{x,y}$ converges, as illustrated by the following example.

Example 2. Let $I := \{-1,1\}$. For all $x\in I$ and $y\in \mathbb{N}$ we define $f_{x,y} := x\cdot y$. Now we have $$\sum_{y=0}^\infty \sum_{x\in I} f_{x,y} = \sum_{y=0}^\infty (-y + y) = \sum_{y=0}^\infty 0 = 0,$$ whereas $$\sum_{x\in I} \sum_{y=0}^\infty f_{x,y} = \left(\sum_{y=0}^\infty -y\right) + \left(\sum_{y=0}^\infty y\right) = -\infty + \infty,$$ which is undefined.

The point is that you still have to be a bit careful, and not blindly rearrange combinations of finite and infinite sums as they come up.

When $I$ is countably infinite, the best tool available is Fubini's theorem from measure theory, as already pointed out by others (João Ramos and A.Γ.). If the double series is not absolutely convergent, you will probably have to examine the outcome on a case-by-case basis.

Like in the case where $I$ was finite, it may happen that only one of the two double sums exists (c.f. example 2 above). However, even if both double series converge, their values might be unequal. Good examples of this have already been given on this website, for instance here and here.