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

  • $\begingroup$ If $\lim_{n\to\infty}f(n)$ diverges, how does the product have any meaning? $\endgroup$ Commented Sep 5, 2016 at 20:14
  • $\begingroup$ @SimpleArt the assumption that $\sum\limits_{n = 1}^\infty \lvert f(n)\rvert$ converges forces $\lim\limits_{n\to \infty} f(n) = 0$. $\endgroup$
    – kobe
    Commented Sep 5, 2016 at 20:15
  • $\begingroup$ Yeah, I know that $\lim_{n \rightarrow \infty} f(n) = 0$, but does it help proving the statement ? Not sure how it is connect to the problem $\endgroup$
    – Both Htob
    Commented Sep 5, 2016 at 20:36
  • $\begingroup$ Use the argumentation of mathworld.wolfram.com/EulerProduct.html and substitute $p_k^{-s}$ by $f(p_k)$. $\endgroup$
    – user90369
    Commented Sep 5, 2016 at 20:44

1 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 $n\rightarrow\infty$.


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