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  "body": "\n## Proof\n\nWe reduce the test to [[Dirichlet's test|0V9ME8GQE7]].\n\n1. Let $\\sum a_n$ be a [[convergent|0CVVKY7GY2]] series with [[partial sums|0NB643K60Q]] $A_N = \\sum_{n \\le N} a_n$ and sum $A$. Let $(b_n)$ be [[monotone|0KN22AAW5R]] and [[bounded|05FQ3VGAR9]].\n2. If $(b_n)$ is increasing, it is bounded above, so it converges by the [[monotone convergence theorem|0FM6N4R0VB]].\n3. If $(b_n)$ is decreasing, it is bounded below, so it converges by the [[decreasing form of that theorem|02V2GYAC8P]].\n4. Let $b$ be the limit of $(b_n)$ and put $d_n = b_n - b$. Then $(d_n)$ is monotone, of the same type as $(b_n)$, and $d_n \\to 0$.\n5. The sequence $(A_N)$ converges, so it is [[bounded|0S4JF619V7]]. That is, there is $K$ with $\\abs{A_N} \\le K$ for all $N$.\n6. By steps 4 and 5, Dirichlet's test applies to the sequences $(d_n)$ and $(a_n)$, after a shift of the index by one. So $\\sum d_n a_n$ converges, with partial sums $D_N = \\sum_{n \\le N} a_n (b_n - b)$.\n7. For each $N$, $\\sum_{n \\le N} a_n b_n = D_N + b A_N$, since $a_n b_n = a_n (b_n - b) + b a_n$.\n8. The sequence $(b A_N)$ converges to $b A$, by the [[product rule for limits|0734RSVF2V]] with the constant sequence $b$.\n9. By the [[sum rule for limits|0RAM4E07NR]], the partial sums $D_N + b A_N$ converge. So $\\sum a_n b_n$ converges.\n\n## Idea\n\nA bounded monotone sequence has a limit $b$. Subtracting $b$ gives a monotone sequence that tends to $0$, which is exactly the hypothesis of Dirichlet's test. The leftover part $b \\sum a_n$ converges because $\\sum a_n$ does.\n",
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