Exercise 8.3

Let ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_j{a}_{\mathit{ij}}=b_i\in [0,\infty ]$. If $\sum <!--\; FUNCTION\; APPLICATION\; -->b_i$ diverges, then the conclusion follows from Theorem 8.3, so suppose $\sum <!--\; FUNCTION\; APPLICATION\; -->b_i$ converges. Let ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_i{a}_{\mathit{ij}}=c_j$, we want to show that $\sum <!--\; FUNCTION\; APPLICATION\; -->c_j=\sum b_i=\infty$. But if $\sum <!--\; FUNCTION\; APPLICATION\; -->c_j$ converges, then we can define the transposed double sequence ${ã}_{\mathit{ij}}=a_{\mathit{ji}}$, and let ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_j{ã}_{\mathit{ij}}=\widetilde{b_i}=c_i$. Since $\sum <!--\; FUNCTION\; APPLICATION\; -->\widetilde{b_i}=\sum c_j$ converges, we can apply Theorem 8.3 to conclude that $\sum <!--\; FUNCTION\; APPLICATION\; -->{\tilde{c}}_i$ converges, where ${\tilde{c}}_i={\sum <!--\; FUNCTION\; APPLICATION\; -->}_j{ã}_{\mathit{ij}}=b_i$. But this contradicts the assumption that $\sum <!--\; FUNCTION\; APPLICATION\; -->b_i$ diverges.