{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\n1. Let $\\sum a_n$ [[converge absolutely|0W3K3QPEWT]] with sum $A$, and let $\\sum b_n$ [[converge|0CVVKY7GY2]] with sum $B$. 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 the three series. Let $S = \\sum_{k=0}^{\\infty} \\abs{a_k}$.\n3. The partial sums of $\\sum \\abs{a_k}$ increase to $S$. So $\\sum_{k=0}^{n} \\abs{a_k} \\le S$ for all $n$, by the [[criterion for nonnegative series|0MX1HZBBXJ]].\n4. Exchange the order of the finite double sum over $0 \\le k \\le n \\le N$, and put $j = n - k$:\n$$C_N = \\sum_{n=0}^{N} \\sum_{k=0}^{n} a_k b_{n-k} = \\sum_{k=0}^{N} a_k \\sum_{j=0}^{N-k} b_j = \\sum_{k=0}^{N} a_k B_{N-k}.$$\n5. Let $\\beta_n = B_n - B$. Then $\\beta_n \\to 0$, and step 4 gives $C_N = A_N B + R_N$ with $R_N = \\sum_{k=0}^{N} a_k \\beta_{N-k}$.\n6. By the [[product rule for limits|0734RSVF2V]], $A_N B \\to AB$. By the [[sum rule for limits|0RAM4E07NR]], it suffices to show $R_N \\to 0$.\n7. The sequence $(\\beta_n)$ converges, so it is [[bounded|0S4JF619V7]]. Fix $K \u003e 0$ with $\\abs{\\beta_n} \\le K$ for all $n$.\n8. Let $\\varepsilon \u003e 0$. By the [[epsilon-N form|09G6KHS6GY]], there is $M \\ge 1$ with $\\abs{\\beta_n} \u003c \\varepsilon / (2S + 2)$ for all $n \\ge M$.\n9. The series $\\sum \\abs{a_k}$ converges. By the [[Cauchy criterion for series|06PVXMY0TH]], there is $L$ with $\\sum_{k=m+1}^{n} \\abs{a_k} \u003c \\varepsilon / (2K)$ for all $n \u003e m \\ge L$.\n10. Let $N \\ge L + M$. Split $R_N$ into the part with $k \\le N - M$ and the part with $N - M \u003c k \\le N$.\n11. In the first part, $N - k \\ge M$, so $\\abs{\\beta_{N-k}} \u003c \\varepsilon / (2S + 2)$. With step 3, the first part has absolute value at most $S \\varepsilon / (2S + 2) \u003c \\varepsilon / 2$.\n12. In the second part, $\\abs{\\beta_{N-k}} \\le K$. With $m = N - M \\ge L$ in step 9, the second part has absolute value at most $K \\sum_{k=N-M+1}^{N} \\abs{a_k} \u003c \\varepsilon / 2$.\n13. By the [[triangle inequality|03HF5PMY5T]], $\\abs{R_N} \u003c \\varepsilon$ for all $N \\ge L + M$. So $R_N \\to 0$.\n14. Hence $C_N \\to AB$. The Cauchy product converges, and its sum is $AB$.\n\n## Idea\n\nRewrite the partial sum of the Cauchy product as $\\sum a_k B_{N-k}$. Replacing each $B_{N-k}$ by $B$ gives $A_N B$, and the error is a weighted sum of the small numbers $B_{N-k} - B$. For most $k$ the error factor is small, and the remaining few indices carry only a small tail of $\\sum \\abs{a_k}$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "40A05"
    ],
    "name": "Proof of Mertens' theorem on Cauchy products",
    "proves": "0BJSWZJG18",
    "sources": [
      {
        "locator": "Convergence and Mertens' theorem",
        "url": "https://en.wikipedia.org/wiki/Cauchy_product"
      }
    ],
    "tag": "6V2VSM8A89",
    "uses": [
      "0W3K3QPEWT",
      "0CVVKY7GY2",
      "0B1K9PVWE9",
      "0NB643K60Q",
      "0S4JF619V7",
      "06PVXMY0TH",
      "0MX1HZBBXJ",
      "09G6KHS6GY",
      "0734RSVF2V",
      "0RAM4E07NR",
      "03HF5PMY5T"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/6V2VSM8A89/",
  "links": [
    {
      "field": "proves",
      "target": "0BJSWZJG18"
    },
    {
      "field": "uses",
      "target": "0W3K3QPEWT"
    },
    {
      "field": "uses",
      "target": "0CVVKY7GY2"
    },
    {
      "field": "uses",
      "target": "0B1K9PVWE9"
    },
    {
      "field": "uses",
      "target": "0NB643K60Q"
    },
    {
      "field": "uses",
      "target": "0S4JF619V7"
    },
    {
      "field": "uses",
      "target": "06PVXMY0TH"
    },
    {
      "field": "uses",
      "target": "0MX1HZBBXJ"
    },
    {
      "field": "uses",
      "target": "09G6KHS6GY"
    },
    {
      "field": "uses",
      "target": "0734RSVF2V"
    },
    {
      "field": "uses",
      "target": "0RAM4E07NR"
    },
    {
      "field": "uses",
      "target": "03HF5PMY5T"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/6V2VSM8A89.md",
  "msc": [
    "40A05"
  ],
  "name": "Proof of Mertens' theorem on Cauchy products",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/6V2/6V2VSM8A89.md",
  "status": null,
  "tag": "6V2VSM8A89",
  "uri": "m2://6V2VSM8A89",
  "url": "https://math2.tamnd.com/wiki/proof-of-mertens-theorem-on-cauchy-products/",
  "writing": {
    "tag": "6V2VSM8A89",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}