Determing sequence from its Dirichlet series 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?
 A: 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).
