(January 2006 #5) Let \(a_{m,n}\ge0\) for \(m,n\in\mathbb N\) and assume that the partial sums \[ \sum_{m=1}^{M}\sum_{n=1}^{N}a_{m,n} \] are bounded above. Prove carefully that \[ \sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}a_{m,n}\right) \quad\text{and}\quad \sum_{n=1}^{\infty}\left(\sum_{m=1}^{\infty}a_{m,n}\right) \] exist and are equal.
Solution
1. The doubly-indexed partial sums are bounded, by assumption; call \(B = \sup_{M,N}\sum_{m=1}^M\sum_{n=1}^N a_{m,n} < \infty\).
2. The inner series exist and the iterated sums exist in \([0,\infty]\).Proof: for each fixed \(m\), the partial sums \(\sum_{n=1}^N a_{m,n}\) are increasing in \(N\) and bounded by \(B\), so \(A_m := \sum_{n=1}^\infty a_{m,n}\) converges with \(A_m \le B\).Then \(\sum_{m=1}^M A_m\) is increasing in \(M\) and bounded by \(B\) (as \(\sum_{m=1}^M A_m = \sup_N\sum_{m=1}^M\sum_{n=1}^N a_{m,n} \le B\)), so \(\sum_{m=1}^\infty A_m\) converges and equals \(\sup_M\sum_{m=1}^M A_m \le B\).Symmetrically \(\sum_{n=1}^\infty\big(\sum_{m=1}^\infty a_{m,n}\big)\) converges.
3. Both iterated sums equal \(\sup_{M,N} S_{M,N}\).Proof: for every \(M, N\), \(S_{M,N} := \sum_{m=1}^M\sum_{n=1}^N a_{m,n} \le \sum_{m=1}^M A_m \le \sup_M\sum_{m=1}^M A_m\), so \(\sup_{M,N}S_{M,N} \le \sum_{m=1}^\infty A_m\).Conversely, for fixed \(M\), \(S_{M,N} \nearrow \sum_{m=1}^M A_m\) as \(N \to \infty\) (monotone convergence), so \(\sum_{m=1}^M A_m = \sup_N S_{M,N} \le \sup_{M,N}S_{M,N}\); taking the sup over \(M\) gives \(\sum_{m=1}^\infty A_m \le \sup_{M,N}S_{M,N}\).Hence equality: \(\sum_{m=1}^\infty\big(\sum_{n=1}^\infty a_{m,n}\big) = \sup_{M,N}S_{M,N}\).The same argument with the roles of \(m\) and \(n\) exchanged gives \(\sum_{n=1}^\infty\big(\sum_{m=1}^\infty a_{m,n}\big) = \sup_{M,N}S_{M,N}\).
4. Q.E.D. Proof: both iterated sums exist and equal the same finite number \(\sup_{M,N}S_{M,N}\).