[proofplan] We rewrite the two absolutely convergent Dirichlet series as ordinary absolutely convergent series with terms $A_a=f(a)a^{-s}$ and $B_b=g(b)b^{-s}$. Absolute convergence makes the double family $(A_aB_b)_{(a,b)\in\mathbb{N}^2}$ summable, with total absolute mass equal to the product of the two absolute sums. We then group this absolutely convergent double series according to the product $ab=n$; the grouped coefficient is exactly the Dirichlet convolution coefficient $(f*g)(n)n^{-s}$. [/proofplan] [step:Reduce the product to an absolutely summable double family] Define the sequences $A:\mathbb{N}\to\mathbb{C}$ and $B:\mathbb{N}\to\mathbb{C}$ by \begin{align*} A(a) &= f(a)a^{-s}, & B(b) &= g(b)b^{-s}. \end{align*} For readability write $A_a:=A(a)$ and $B_b:=B(b)$. By hypothesis, \begin{align*} \sum_{a=1}^{\infty}|A_a|<\infty, \qquad \sum_{b=1}^{\infty}|B_b|<\infty. \end{align*} Define the double-indexed family $C:\mathbb{N}^2\to\mathbb{C}$ by \begin{align*} C(a,b)=A_aB_b. \end{align*} Then \begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|C(a,b)| &= \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|A_a||B_b| \\ &= \left(\sum_{a=1}^{\infty}|A_a|\right) \left(\sum_{b=1}^{\infty}|B_b|\right) <\infty. \end{align*} Thus the double series over $\mathbb{N}^2$ is absolutely convergent. For each $N\in\mathbb{N}$, define the rectangular partial sums \begin{align*} P_N=\sum_{a=1}^{N}A_a, \qquad Q_N=\sum_{b=1}^{N}B_b. \end{align*} Then \begin{align*} P_NQ_N = \sum_{a=1}^{N}\sum_{b=1}^{N}A_aB_b = \sum_{a=1}^{N}\sum_{b=1}^{N}C(a,b). \end{align*} Since $P_N\to F(s)$ and $Q_N\to G(s)$, the left-hand side converges to $F(s)G(s)$. Since the double family $C$ is absolutely summable, the rectangular sums on the right converge to the full double sum. Hence \begin{align*} F(s)G(s)=\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b. \end{align*} [guided] The point of this step is to separate the analytic issue from the arithmetic one. Define \begin{align*} A:\mathbb{N}&\to\mathbb{C}, & a&\mapsto f(a)a^{-s},\\ B:\mathbb{N}&\to\mathbb{C}, & b&\mapsto g(b)b^{-s}. \end{align*} The hypotheses say exactly that the series $\sum_{a=1}^{\infty}A_a$ and $\sum_{b=1}^{\infty}B_b$ converge absolutely. Now define \begin{align*} C:\mathbb{N}^2&\to\mathbb{C}, & (a,b)&\mapsto A_aB_b. \end{align*} We check absolute summability of this double family directly: \begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|C(a,b)| &= \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|A_a||B_b| \\ &= \sum_{a=1}^{\infty}|A_a|\left(\sum_{b=1}^{\infty}|B_b|\right) \\ &= \left(\sum_{a=1}^{\infty}|A_a|\right) \left(\sum_{b=1}^{\infty}|B_b|\right) <\infty. \end{align*} This is the exact place where absolute convergence is used. It guarantees that the double series can be summed without dependence on the order of summation. For $N\in\mathbb{N}$, let \begin{align*} P_N=\sum_{a=1}^{N}A_a, \qquad Q_N=\sum_{b=1}^{N}B_b. \end{align*} Then finite distributivity gives \begin{align*} P_NQ_N = \left(\sum_{a=1}^{N}A_a\right) \left(\sum_{b=1}^{N}B_b\right) = \sum_{a=1}^{N}\sum_{b=1}^{N}A_aB_b. \end{align*} Because $P_N\to F(s)$ and $Q_N\to G(s)$, the products $P_NQ_N$ converge to $F(s)G(s)$. Since the family $(A_aB_b)_{(a,b)\in\mathbb{N}^2}$ is absolutely summable, the rectangular finite sums converge to the full double sum. Therefore \begin{align*} F(s)G(s)=\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b. \end{align*} [/guided] [/step] [step:Group the double series by the product $ab$] For each $n\in\mathbb{N}$, define \begin{align*} H_n=\sum_{\substack{(a,b)\in\mathbb{N}^2\\ab=n}}A_aB_b. \end{align*} The set of pairs $(a,b)\in\mathbb{N}^2$ with $ab=n$ is finite, so $H_n$ is well-defined. Moreover, \begin{align*} \sum_{n=1}^{\infty}|H_n| &\leq \sum_{n=1}^{\infty} \sum_{\substack{(a,b)\in\mathbb{N}^2\\ab=n}} |A_aB_b| \\ &= \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|A_aB_b| <\infty. \end{align*} Thus $\sum_{n=1}^{\infty}H_n$ is absolutely convergent. For each $N\in\mathbb{N}$, define the finite set \begin{align*} E_N=\{(a,b)\in\mathbb{N}^2:ab\leq N\}. \end{align*} Then \begin{align*} \sum_{n=1}^{N}H_n = \sum_{(a,b)\in E_N}A_aB_b. \end{align*} The sets $E_N$ increase to $\mathbb{N}^2$. Since the double family $(A_aB_b)$ is absolutely summable, passing to the limit gives \begin{align*} \sum_{n=1}^{\infty}H_n = \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b. \end{align*} Combining this with the previous step, \begin{align*} F(s)G(s)=\sum_{n=1}^{\infty}H_n. \end{align*} [guided] Now we reorganize the absolutely convergent double series by the integer value of the product $ab$. For each $n\in\mathbb{N}$, define \begin{align*} H_n=\sum_{\substack{(a,b)\in\mathbb{N}^2\\ab=n}}A_aB_b. \end{align*} This is a finite sum because $ab=n$ implies $a$ is a positive divisor of $n$, and then $b=n/a$ is determined. We first prove that the grouped series is absolutely convergent. By the triangle inequality for each finite divisor sum, \begin{align*} |H_n| \leq \sum_{\substack{(a,b)\in\mathbb{N}^2\\ab=n}}|A_aB_b|. \end{align*} Summing over $n$ gives \begin{align*} \sum_{n=1}^{\infty}|H_n| &\leq \sum_{n=1}^{\infty} \sum_{\substack{(a,b)\in\mathbb{N}^2\\ab=n}} |A_aB_b| \\ &= \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|A_aB_b| <\infty. \end{align*} The equality holds because each pair $(a,b)\in\mathbb{N}^2$ appears exactly once, namely in the group indexed by $n=ab$. To identify the value of the grouped sum, define \begin{align*} E_N=\{(a,b)\in\mathbb{N}^2:ab\leq N\}. \end{align*} This is a finite set for each $N$. By the definition of $H_n$, \begin{align*} \sum_{n=1}^{N}H_n = \sum_{(a,b)\in E_N}A_aB_b. \end{align*} The sets $E_N$ exhaust $\mathbb{N}^2$: for every pair $(a,b)\in\mathbb{N}^2$, the pair belongs to $E_N$ whenever $N\geq ab$. Since $\sum_{a,b}|A_aB_b|<\infty$, the contribution from the complement $\mathbb{N}^2\setminus E_N$ tends to $0$. Therefore \begin{align*} \sum_{n=1}^{\infty}H_n = \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b. \end{align*} Together with the identity from the previous step, this gives \begin{align*} F(s)G(s)=\sum_{n=1}^{\infty}H_n. \end{align*} [/guided] [/step] [step:Identify the grouped coefficients with the Dirichlet convolution] Fix $n\in\mathbb{N}$. The map from positive divisors of $n$ to pairs $(a,b)\in\mathbb{N}^2$ satisfying $ab=n$, \begin{align*} d \mapsto (d,n/d), \end{align*} is a bijection. Hence \begin{align*} H_n &= \sum_{\substack{(a,b)\in\mathbb{N}^2\\ab=n}}A_aB_b \\ &= \sum_{d\mid n}A_dB_{n/d} \\ &= \sum_{d\mid n}f(d)d^{-s}g(n/d)(n/d)^{-s}. \end{align*} Since $d(n/d)=n$ and the powers are taken using the real logarithm on positive integers, \begin{align*} d^{-s}(n/d)^{-s}=n^{-s}. \end{align*} Therefore \begin{align*} H_n &= \left(\sum_{d\mid n}f(d)g(n/d)\right)n^{-s} \\ &= (f*g)(n)n^{-s}. \end{align*} Substituting this identity into the grouped series gives \begin{align*} F(s)G(s)=\sum_{n=1}^{\infty}(f*g)(n)n^{-s}. \end{align*} The absolute convergence of this Dirichlet series follows from the already proved convergence of $\sum_{n=1}^{\infty}|H_n|$. This proves the theorem. [/step]