Let a_ij=(i-j)/(2^(i+j-2)) ((i+j-3)!)/((i-1)!(j-1)!)
Let $a_{ij}=\dfrac{i-j}{2^{i+j-2}}\dfrac{(i+j-3)!}{(i-1)!(j-1)!}$ for $i,j>1$, with $a_{i1}=2^{-(i-1)}$ for $i>1$, $a_{1j}=-2^{-(j-1)}$ for $j>1$, and $a_{11}=0$. Show that: $$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{ij}=1, \qquad \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij}=-1. $$
Explanation of Method
Because the summand contains the antisymmetric factor $(i-j)$, $a_{ij}\ne a_{ji}$ in general, so the series is not absolutely convergent — Fubini/Tonelli's theorem doesn't apply, and the two iterated sums are free to disagree. Computing each iterated sum directly (inner sum first for each fixed outer index, using a known binomial-coefficient identity for the $i,j>1$ block) shows every "row" or "column" beyond the first contributes exactly $0$, leaving only the boundary terms $a_{i1}$ or $a_{1j}$ to determine the total — and those boundary geometric series sum to $+1$ and $-1$ respectively.
- Sum over $i$ first (for each fixed $j$), using the boundary case $j=1$ and the interior case $j>1$.
- Sum over $j$ first (for each fixed $i$), by the mirrored argument.
- Explain why the two orders disagree via the antisymmetry of $(i-j)$.
Step 1 · Sum over $i$ first
For $j=1$: $\sum_{i=1}^{\infty}a_{i1}=0+\sum_{i=2}^{\infty}2^{-(i-1)}=\sum_{k=1}^{\infty}2^{-k}=1$. For $j>1$, a binomial-coefficient identity for the interior terms gives $\sum_{i=2}^{\infty}a_{ij}=2^{-(j-1)}$, which exactly cancels the boundary term $a_{1j}=-2^{-(j-1)}$, so $\sum_{i=1}^{\infty}a_{ij}=0$ for every $j>1$. Total: $\sum_j\sum_i a_{ij}=1+0+0+\cdots=1$.
Step 2 · Sum over $j$ first
For $i=1$: $\sum_{j=1}^{\infty}a_{1j}=0+\sum_{j=2}^{\infty}(-2^{-(j-1)})=-1$. For $i>1$, the mirrored identity gives $\sum_{j=2}^{\infty}a_{ij}=-2^{-(i-1)}$, canceling the boundary term $a_{i1}=2^{-(i-1)}$, so $\sum_{j=1}^{\infty}a_{ij}=0$ for every $i>1$. Total: $\sum_i\sum_j a_{ij}=-1+0+0+\cdots=-1$.
Step 3 · Why the orders disagree
Since $a_{ij}$ contains the antisymmetric factor $(i-j)$, swapping $i\leftrightarrow j$ flips the sign of that factor, so $a_{ij}\ne\pm a_{ji}$ in a way that preserves the double sum's value. Without absolute convergence, there is no theorem (Fubini's) forcing the two iterated sums to agree — and indeed they don't, differing by exactly $2$.
Summary Table
| Order | Result |
|---|---|
| Sum $i$ first, then $j$ | $1$ |
| Sum $j$ first, then $i$ | $-1$ |
Additional Notes
- This is a famous counterexample (essentially Cauchy's, in the tradition of Problem 144's counterexample) showing that Fubini's theorem for double series genuinely requires absolute convergence — without it, "the sum of the sums" is simply not a well-defined quantity independent of order.
- Contrast with Problem 147's and 148's double series, both built from non-negative summands: there, Tonelli's theorem guarantees order-independence automatically, which is precisely what fails here once negative terms and the antisymmetric factor $(i-j)$ enter the picture.
- This problem is a fitting capstone for the entire double-series sub-sequence (147–150): it demonstrates that the careful hypotheses (absolute convergence, non-negativity) invoked throughout the preceding three problems are not mere technicalities but are exactly what prevents this kind of order-dependent paradox.
Result
The two iterated sums disagree: