---
tag: 6V2VSM8A89
kind: proof
name: Proof of Mertens' theorem on Cauchy products
msc: [40A05]
proves: 0BJSWZJG18
uses: [0W3K3QPEWT, 0CVVKY7GY2, 0B1K9PVWE9, 0NB643K60Q, 0S4JF619V7, 06PVXMY0TH, 0MX1HZBBXJ, 09G6KHS6GY, 0734RSVF2V, 0RAM4E07NR, 03HF5PMY5T]
sources:
  - url: https://en.wikipedia.org/wiki/Cauchy_product
    locator: Convergence and Mertens' theorem
---

## Proof

1. 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]].
2. Let $A_N$, $B_N$, $C_N$ be the [[partial sums|0NB643K60Q]] of the three series. Let $S = \sum_{k=0}^{\infty} \abs{a_k}$.
3. 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]].
4. Exchange the order of the finite double sum over $0 \le k \le n \le N$, and put $j = n - k$:
$$C_N = \sum_{n=0}^{N} \sum_{k=0}^{n} a_k b_{n-k} = \sum_{k=0}^{N} a_k \sum_{j=0}^{N-k} b_j = \sum_{k=0}^{N} a_k B_{N-k}.$$
5. Let $\beta_n = B_n - B$. Then $\beta_n \to 0$, and step 4 gives $C_N = A_N B + R_N$ with $R_N = \sum_{k=0}^{N} a_k \beta_{N-k}$.
6. By the [[product rule for limits|0734RSVF2V]], $A_N B \to AB$. By the [[sum rule for limits|0RAM4E07NR]], it suffices to show $R_N \to 0$.
7. The sequence $(\beta_n)$ converges, so it is [[bounded|0S4JF619V7]]. Fix $K > 0$ with $\abs{\beta_n} \le K$ for all $n$.
8. Let $\varepsilon > 0$. By the [[epsilon-N form|09G6KHS6GY]], there is $M \ge 1$ with $\abs{\beta_n} < \varepsilon / (2S + 2)$ for all $n \ge M$.
9. The series $\sum \abs{a_k}$ converges. By the [[Cauchy criterion for series|06PVXMY0TH]], there is $L$ with $\sum_{k=m+1}^{n} \abs{a_k} < \varepsilon / (2K)$ for all $n > m \ge L$.
10. Let $N \ge L + M$. Split $R_N$ into the part with $k \le N - M$ and the part with $N - M < k \le N$.
11. In the first part, $N - k \ge M$, so $\abs{\beta_{N-k}} < \varepsilon / (2S + 2)$. With step 3, the first part has absolute value at most $S \varepsilon / (2S + 2) < \varepsilon / 2$.
12. 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 $K \sum_{k=N-M+1}^{N} \abs{a_k} < \varepsilon / 2$.
13. By the [[triangle inequality|03HF5PMY5T]], $\abs{R_N} < \varepsilon$ for all $N \ge L + M$. So $R_N \to 0$.
14. Hence $C_N \to AB$. The Cauchy product converges, and its sum is $AB$.

## Idea

Rewrite 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 $\sum \abs{a_k}$.