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  "body": "\n## Proof\n\nWe use summation by parts and the Cauchy criterion.\n\n1. Let $(a_n)_{n \\ge 1}$ be [[monotone|0KN22AAW5R]] with $a_n \\to 0$. Let $(b_n)_{n \\ge 1}$ be real with $\\abs{B_n} \\le M$ for all $n$, where $B_n = \\sum_{k=1}^{n} b_k$ is the $n$-th [[partial sum|0NB643K60Q]].\n2. Put $B_0 = 0$. Then $b_k = B_k - B_{k-1}$ for all $k \\ge 1$, and $\\abs{B_k} \\le M$ for all $k \\ge 0$.\n3. Let $n \u003e m \\ge 0$. Then\n$$\\sum_{k=m+1}^{n} a_k b_k = \\sum_{k=m+1}^{n} a_k B_k - \\sum_{k=m+1}^{n} a_k B_{k-1} = \\sum_{k=m+1}^{n} a_k B_k - \\sum_{k=m}^{n-1} a_{k+1} B_k.$$\n4. Collecting the terms with the same $B_k$ in step 3 gives the summation by parts formula\n$$\\sum_{k=m+1}^{n} a_k b_k = a_n B_n - a_{m+1} B_m + \\sum_{k=m+1}^{n-1} (a_k - a_{k+1}) B_k.$$\n5. Since $(a_n)$ is monotone, the differences $a_k - a_{k+1}$ all have the same sign. So $\\sum_{k=m+1}^{n-1} \\abs{a_k - a_{k+1}} = \\abs{\\sum_{k=m+1}^{n-1} (a_k - a_{k+1})}$.\n6. The last sum telescopes to $a_{m+1} - a_n$. So $\\sum_{k=m+1}^{n-1} \\abs{a_k - a_{k+1}} = \\abs{a_{m+1} - a_n} \\le \\abs{a_{m+1}} + \\abs{a_n}$.\n7. By step 4, the [[triangle inequality|03HF5PMY5T]], $\\abs{B_k} \\le M$ and step 6,\n$$\\abs{\\sum_{k=m+1}^{n} a_k b_k} \\le M \\abs{a_n} + M \\abs{a_{m+1}} + M \\left( \\abs{a_{m+1}} + \\abs{a_n} \\right) = 2M \\left( \\abs{a_n} + \\abs{a_{m+1}} \\right).$$\n8. Let $\\varepsilon \u003e 0$. Since $a_n \\to 0$, the [[epsilon-N form of the limit|09G6KHS6GY]] gives $N \\ge 1$ with $\\abs{a_k} \u003c \\varepsilon / (4M + 1)$ for all $k \\ge N$.\n9. For $n \u003e m \\ge N$, both $n$ and $m + 1$ are at least $N$. So step 7 gives $\\abs{\\sum_{k=m+1}^{n} a_k b_k} \\le 4M \\varepsilon / (4M + 1) \u003c \\varepsilon$.\n10. The Cauchy criterion is stated for series indexed from $0$. Adding the term $0$ at index $0$ leaves the partial sums unchanged.\n11. By step 9 and the [[Cauchy criterion for series|06PVXMY0TH]], the series $\\sum_{n=1}^{\\infty} a_n b_n$ converges.\n\n## Idea\n\nSummation by parts moves the differences from $b_n$ to $a_n$. The partial sums $B_n$ are bounded, and the differences $a_k - a_{k+1}$ have one sign, so their absolute values telescope. Every block of the series is then bounded by a multiple of a few terms $\\abs{a_n}$, which tend to $0$.\n",
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