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  "body": "\n## Proof\n\nWe compare the partial sums of the Cauchy product with products of partial sums.\n\n1. Let $\\sum a_n$ and $\\sum b_n$ [[converge absolutely|0W3K3QPEWT]], with sums $A$ and $B$. Let $c_n = \\sum_{k=0}^{n} a_k b_{n-k}$ define the [[Cauchy product|0B1K9PVWE9]].\n2. Let $A_N$, $B_N$, $C_N$ be the [[partial sums|0NB643K60Q]] of $\\sum a_n$, $\\sum b_n$, $\\sum c_n$. Let $P_N = \\sum_{n \\le N} \\abs{a_n}$ and $Q_N = \\sum_{n \\le N} \\abs{b_n}$.\n3. Let $P = \\sum \\abs{a_n}$ and $Q = \\sum \\abs{b_n}$. The terms are nonnegative, so $P_N \\le P$ and $Q_N \\le Q$ by the [[criterion for nonnegative series|0MX1HZBBXJ]].\n4. The term $c_n$ is the sum of $a_i b_j$ over the pairs $(i, j)$ of natural numbers with $i + j = n$. So $C_N$ is the sum of $a_i b_j$ over the set $T_N$ of pairs with $i + j \\le N$.\n5. Each pair in $T_N$ has $i \\le N$ and $j \\le N$. By the [[triangle inequality|03HF5PMY5T]],\n$$\\sum_{n=0}^{N} \\abs{c_n} \\le \\sum_{(i,j) \\in T_N} \\abs{a_i} \\abs{b_j} \\le \\sum_{i=0}^{N} \\sum_{j=0}^{N} \\abs{a_i} \\abs{b_j} = P_N Q_N \\le P Q.$$\n6. So the partial sums of $\\sum \\abs{c_n}$ are bounded above by $PQ$. By the criterion for nonnegative series, $\\sum \\abs{c_n}$ converges. Hence $\\sum c_n$ converges absolutely.\n7. Let $U_N$ be the set of pairs $(i, j)$ with $i \\le N$, $j \\le N$ and $i + j \u003e N$. Then $A_N B_N = \\sum_{i, j \\le N} a_i b_j = C_N + \\sum_{(i,j) \\in U_N} a_i b_j$.\n8. Let $E_N = \\sum_{(i,j) \\in U_N} \\abs{a_i} \\abs{b_j}$. By step 7 and the triangle inequality, $\\abs{A_N B_N - C_N} \\le E_N$.\n9. In the same way as step 7, $P_N Q_N = \\sum_{(i,j) \\in T_N} \\abs{a_i} \\abs{b_j} + E_N$.\n10. Let $h = \\floor{N/2}$. If $i \\le h$ and $j \\le h$, then $i + j \\le N$. So $\\sum_{(i,j) \\in T_N} \\abs{a_i} \\abs{b_j} \\ge P_h Q_h$.\n11. By steps 9 and 10, $0 \\le E_N \\le P_N Q_N - P_{\\floor{N/2}} Q_{\\floor{N/2}}$.\n12. By the [[product rule for limits|0734RSVF2V]], $P_N Q_N \\to PQ$. Let $\\varepsilon \u003e 0$, and take $K$ with $\\abs{P_k Q_k - PQ} \u003c \\varepsilon$ for $k \\ge K$, by the [[epsilon-N form|09G6KHS6GY]].\n13. For $N \\ge 2K$ we have $\\floor{N/2} \\ge K$. So $P_{\\floor{N/2}} Q_{\\floor{N/2}} \\to PQ$ as $N \\to \\infty$.\n14. By the [[sum rule for limits|0RAM4E07NR]], the right side of step 11 tends to $0$. By the [[squeeze theorem|0XMF00T4XQ]], $E_N \\to 0$.\n15. By step 8, $-E_N \\le A_N B_N - C_N \\le E_N$. The squeeze theorem gives $A_N B_N - C_N \\to 0$.\n16. By the product rule, $A_N B_N \\to AB$. So $C_N = A_N B_N - (A_N B_N - C_N) \\to AB - 0 = AB$ by the sum rule.\n17. Hence $\\sum c_n$ converges absolutely, and its sum is $AB$.\n\n## Idea\n\nThe product $A_N B_N$ sums $a_i b_j$ over a square of indices, while $C_N$ sums over the triangle $i + j \\le N$ inside it. The missing corner is controlled by the same corner for the absolute values. That corner lies between the squares of sides $N/2$ and $N$, so it tends to $0$.\n",
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