example of a multiplicative function Can anybody give me an example of a multiplicative function $f$ such that
$$\prod_p \sum_{k=0}^\infty f(p^k)$$
converges absolutely and such that
$$\sum_{n=1}^\infty f(n)$$
diverges?
 A: The product $\Pi_p(1+a_p)$ converges absolutely iff the sum $\sum_p a_p$ does. 
A multiplicative function may be defined as $f(1)=1$ and then for each prime $p$ and power $k \ge 1$ the values of $f(p^k)$ may be assigned arbitrarily, and then $f$ itself may be defined by extending from prime powers using the multiplicativity.
So suppose we define for each $p$ that $f(p)=1$, while also choosing the values of $f$ at higher powers of $p$ in such a way that
$$a_p \equiv f(p)+f(p^2)+ \cdots = \frac{1}{p^2}.$$
[This will entail negative values for some of the higher $f(p^k)$, but that has no effect on the sum of all the terms being $1/p^2$, provided we choose them right.]
Then the product $\Pi_p(1+a_p)$ has its $a_p=\frac{1}{p^2}$, and so converges absolutely since the sum of reciprocal squares of primes does.
However the sum $\sum_n f(n)$ diverges, since if it converges then its $n$th term must approach zero, yet for any squarefree number $n$ we have $f(n)=1.$
A: Wherever $\sum_n f(n)$ converges also $\prod_p\sum_k f(p^k)$ does and vice versa.
I emphasize that this is true only inside the convergence region.
This comes from the fact that $f$ is multiplicative and from the fundamental theorem of arithmetic.
So the answer is: no.
