Determing sequence from its Dirichlet series

Question

Suppose I know the Dirchlet series $$\sum_{n=1}^{\infty} \frac{f(n)}{n^s} = \frac{\zeta(s)}{\zeta(3s)},$$ where $\zeta(s)$ is the usual Riemann zeta function.

My question is - is there a way to determine $f(n)$ from this information? If so, how?

Answer 1

Yes: \begin{align} \sum_{n = 1}^\infty \frac{f(n)}{n^s} & = \frac{\zeta(s)}{\zeta(3s)}\\ & = \prod_p \frac{1 - p^{-3s}}{1 - p^{-s}} &&\text{by the Euler product for $\zeta$ where the $p$s are all primes}\\ & = \prod_p \left(1 + \frac{1}{p^s} + \frac{1}{p^{2s}} \right) &&\text{by the identity $ 1 - x^3 = (1 - x)(1 + x + x^2)$} \end{align} Comparing with the Euler product formula, we see that $f$ is a multiplicative function and that $$f(p^\alpha) = \begin{cases} 1 & \text{if $\alpha = 1$ or $2$,} \\ 0 & \text{if $\alpha = 3, 4, 5, \ldots$.} \end{cases}$$

Because a multiplicative function is completely determined by its values at the powers of prime numbers, we can restate the function as $$f(n) = \begin{cases} 1 & \text{if 1 is the largest cube that divides $n$,} \\ 0 & \text{otherwise.} \end{cases}$$

The Dirichlet series for $\zeta(s)$ and $\zeta(3s)$ both converge absolutely for $s > 1$, so that definition of $f(n)$ is the only answer when $s > 1$ (see Theorem 4.8 at http://www.math.illinois.edu/~ajh/ant/main4.pdf).