Summation formula for multiplicative function

Question

Assume $f$ is multiplicative. Assume further that $$\sum_{n=1}^\infty |f(n)|$$ converges. Then show that $$\sum_{n=1}^\infty f(n)= \prod_{p \ \mbox{prime}} (1 + f(p) + f(p^2) + ...).$$

Proof. Let $n = p_{1}^{a_1}p_{2}^{a_2}...p_{k}^{a_k}. $ Then $$f(n) = f(p_1)^{a_1}...f(p_k)^{a_k} = f(p_1)^{a_1}...f(p_k)^{a_k}(1)(1)... $$ Since we have all $n \in \mathbb{N}$ and the left hand side product provides any combination of product of primes without repetition (by the uniqueness of prime power facterization; Fundamantal theorem of Arithmetic), they are intuitively equals.

However, I suppose that I need to use $\delta-\epsilon$ notion which I am not sure how to do it rigorously

Answer 1

The equality is true in the formal ring of power series:

$$\sum_{n=1}^\infty f(n)x^n=\prod_{p \ \mbox{prime}}(1+f(p)x+f(p^2)x^2+\cdots), $$

To show convergence at $x=1$, you can instead look at the product:

$|\prod_{p \ \mbox{prime}, \ p\leq n}(1+f(p)+f(p^2)+\cdots)|\leq \sum_{i=1}^{n^n}|f(i)|.$

Now take $n\rightarrow\infty$.