Conditional Double Series with Different Iterated Sums

When the Order of Summation Matters

A double series can have well-defined row sums and column sums without those two iterated sums agreeing. The distinction is important: a theorem guaranteeing agreement requires absolute summability, not merely the existence of each iterated sum. We examine an array for which each row and each column has only finitely many nonzero entries, yet summing rows first gives a different answer from summing columns first.

Let \(\mathbb{N}=\{1,2,\ldots\}\), equipped with the sigma-algebra \(\mathcal{P}(\mathbb{N})\) of all subsets and counting measure \(\#\). This measure space is sigma-finite and complete, but not finite. The product space has counting measure on \(\mathbb{N}\times\mathbb{N}\). Consider the real-valued function \(a:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\) defined by

a_{mn}= \begin{cases} 1,&n=m,\\ -1,&n=m+1,\\ 0,&\text{otherwise}. \end{cases}

The entries \(1\) lie on the diagonal; the entries \(-1\) lie immediately to its right. Since every subset of \(\mathbb{N}\times\mathbb{N}\) is measurable, \(a\) is measurable. Integration with respect to counting measure is summation, so the row-first and column-first expressions are the two iterated integrals of \(a\). We will calculate them directly; the integral of \(a\) over the whole product space is not defined as a signed Lebesgue integral, as we will verify below.

The Two Iterated Sums

Definition: For a real array \((a_{mn})\), its row-first iterated sum is \(\displaystyle\sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}a_{mn}\right)\), provided each inner row sum and the outer sum are defined. Its column-first iterated sum is \(\displaystyle\sum_{n=1}^{\infty}\left(\sum_{m=1}^{\infty}a_{mn}\right)\), provided each inner column sum and the outer sum are defined. These are ordered procedures; neither notation by itself asserts absolute summability or that the two results agree.

Worked Example: Summing the Rows and the Columns

For fixed \(m\), the only nonzero entries in row \(m\) are \(a_{m,m}=1\) and \(a_{m,m+1}=-1\). Thus its sum is \(1+(-1)=0\). Consequently every outer term in the row-first sum is zero, and

\sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}a_{mn}\right) =\sum_{m=1}^{\infty}0=0.

Now fix a column \(n\). If \(n=1\), the only nonzero entry in that column is \(a_{1,1}=1\), so the column sum is \(1\). If \(n\ge2\), its nonzero entries are \(a_{n-1,n}=-1\) and \(a_{n,n}=1\), giving column sum \((-1)+1=0\). Therefore

\sum_{n=1}^{\infty}\left(\sum_{m=1}^{\infty}a_{mn}\right) =1+\sum_{n=2}^{\infty}0=1.

Check: the row-first outer series is \(0+0+\cdots\), hence equals \(0\). The column-first outer series is \(1+0+0+\cdots\), hence equals \(1\). Both procedures are well-defined, but their answers differ.

The different answers are not caused by an infinite inner row or column sum: every such sum here has at most two nonzero terms. The distinction arises when the finite cancellations are assembled across infinitely many rows or columns. The hypothesis that would guarantee agreement is absolute summability of the whole array.

Theorem (agreement under absolute summability): Let \((a_{mn})\) be a real array on \(\mathbb{N}\times\mathbb{N}\). If \(\displaystyle\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|a_{mn}|<\infty\), then both iterated sums exist as finite real numbers and are equal. This is the countable-sum case of Fubini's theorem for integrable functions on sigma-finite measure spaces.

Proof. Write \(A=\sum_{m,n}|a_{mn}|<\infty\). By Tonelli's theorem for nonnegative series, \(\sum_m\sum_n|a_{mn}|=A\) and \(\sum_n\sum_m|a_{mn}|=A\). In particular, every row and column is absolutely summable. Define \(r_m=\sum_n a_{mn}\) and \(c_n=\sum_m a_{mn}\). The triangle inequality gives \(|r_m|\le\sum_n|a_{mn}|\) and \(|c_n|\le\sum_m|a_{mn}|\). Summing these inequalities shows that \(\sum_m|r_m|\le A\) and \(\sum_n|c_n|\le A\), so both iterated sums converge absolutely.

For each positive integer \(K\), let \(S_K=\sum_{m=1}^K\sum_{n=1}^K a_{mn}\). Absolute convergence implies that the total absolute mass outside the square \(\{1,\ldots,K\}^2\) tends to zero as \(K\to\infty\). The difference between \(\sum_m r_m\) and \(S_K\) consists only of terms outside that square; by the triangle inequality its absolute value is at most that outside absolute mass. The same bound applies to \(\sum_n c_n-S_K\). Thus, for every \(\varepsilon>0\), choose \(K\) so that the outside absolute mass is less than \(\varepsilon\). Then \(|\sum_m r_m-S_K|<\varepsilon\) and \(|\sum_n c_n-S_K|<\varepsilon\), giving \(|\sum_m r_m-\sum_n c_n|<2\varepsilon\). Since this holds for every \(\varepsilon>0\), the two sums are equal. \(\square\)

Rectangular Partial Sums Reveal the Dependence

A finite rectangle gives an unambiguous finite sum. However, different ways of expanding rectangles can yield different limits. For positive integers \(M,N\), define \(S_{M,N}=\sum_{m=1}^M\sum_{n=1}^N a_{mn}\). There are \(\min(M,N)\) diagonal entries equal to \(1\). An entry \(-1\) is included exactly when \(m\le M\) and \(m+1\le N\), so there are \(\min(M,N-1)\) such entries. Therefore

S_{M,N}=\min(M,N)-\min(M,N-1).

Worked Example: Two Expanding Families of Rectangles

For the square \(M=N=K\), the formula gives \(S_{K,K}=K-(K-1)=1\) for every \(K\ge1\). Thus square partial sums tend to \(1\).

Now take \(M=K-1\) and \(N=K\), for \(K\ge2\). There are \(K-1\) diagonal entries and \(K-1\) entries equal to \(-1\), so

S_{K-1,K}=(K-1)-(K-1)=0.

As \(K\to\infty\), both side lengths of these rectangles tend to infinity, but their sums remain \(0\). Check: for \(K=4\), the square has \(4\) diagonal terms and \(3\) negative terms, giving \(4-3=1\); the \(3\)-by-\(4\) rectangle has \(3\) of each, giving \(3-3=0\). The two expanding shapes therefore have different limits.

This calculation helps explain why “the sum of all entries” cannot be assigned a shape-independent value by simply taking larger finite pieces. A particular ordering or family of finite sets can produce a limit, but another can produce a different one. The iterated sums are specific choices of order, not a license to rearrange terms.

Why Absolute Summability Fails

Worked Example: Positive, Negative, and Absolute Mass

On the product space with counting measure, the integral of a nonnegative function is the sum of its values, possibly \(+\infty\). The positive entries of \(a\) occur at \((m,m)\) for every \(m\in\mathbb{N}\). Hence the positive part \(a^+=\max(a,0)\) has sum \(1+1+\cdots=+\infty\). The negative part \(a^-=\max(-a,0)\) equals \(1\) at every \((m,m+1)\), so its sum is also \(1+1+\cdots=+\infty\). In particular,

\int a^+\,d(\#\times\#)=+\infty, \qquad \int a^-\,d(\#\times\#)=+\infty, \qquad \int |a|\,d(\#\times\#)=+\infty.

The last equality follows because \(|a|=1\) at every diagonal and next-diagonal position, and there are infinitely many of each. Equivalently, summing row by row gives \(\sum_m\sum_n|a_{mn}|=\sum_m(1+1)=+\infty\). Thus \(a\) is not absolutely summable and is not integrable on the product space. Since both positive and negative integrals are infinite, its signed Lebesgue integral is undefined, rather than a finite number obtained by cancellation.

Check: summing the absolute values in the first \(K\) rows gives \(2K\), which tends to \(+\infty\). This agrees with the separate calculation that both the positive and negative parts have infinite sum.

A Common Mistake: Cancelling Terms Before Choosing an Order

A tempting argument is to pair each \(+1\) on the diagonal with a \(-1\) immediately to its right and conclude that the total is \(0\). That pairing gives the row-first answer, but it does not establish an order-independent sum. Summing by columns instead leaves the unpaired \(+1\) in column \(1\), giving \(1\). The different rectangular sums above confirm that this is not just a notational discrepancy.

The valid cancellation theorem is the absolute-summability result proved above: when the sum of the absolute values is finite, both iterated sums agree. Here that hypothesis fails because the positive and negative masses are each infinite. It is therefore essential to state the order of summation and to avoid rearranging or regrouping terms without a theorem that permits it.

Check Your Understanding

Use the definitions of the two iterated sums and keep track of which index is summed first.

For the array in this tutorial, compute the sum of column \(n=4\) and explain why column \(n=1\) is exceptional.
Evaluate \(S_{M,N}\) when \(M=5\) and \(N=6\), by counting the positive and negative entries in that rectangle.
Show that square partial sums equal \(1\) for every positive side length, even though the row-first iterated sum equals \(0\).
Why does the absolute-summability theorem not apply to this array? Give both the positive-part and negative-part sums.
State the hypothesis in the agreement theorem that rules out different iterated sums for a real array on \(\mathbb{N}\times\mathbb{N}\).