(absolute) Convergence of a series I want to prove that the following series is convergent for $x>0$:
$$ \sum_{n=1}^\infty \left( \prod_{p\mid n} \frac{1}{p-1}\right) n^{-x} $$
I tried to estimate the product but I didn't get so far...
 A: I'm sure this proof could be tightened a bit, but it works. Consider every subset $S$ of $\mathbb{P}$, the prime numbers, and the corresponding set
$$G(S)=\left\{\prod_{p\in S}p^{f(p)}:f:S\rightarrow\mathbb{N}>0\right\}$$
that is, the set of $n$ such that $S$ is the set of primes dividing it. We can show your the sum is now equal to
$$\sum_{\substack{S\subseteq\mathbb{P}\\S\text{ is finite}}}\left(\prod_{p\in S}\frac{1}{p-1}\right)\cdot\left(\sum_{n\in G(S)}n^{-x}\right)$$
where the only particularly mysterious term is the sum over $G(S)$. However, since $G(S\cup\{p\})$ can be written as $pG(S)\cup p^2G(S)\cup p^3G(S)\ldots$ and exponents distribute over multiplication, we get
$$\sum_{n\in G(S\cup\{p\})}n^{-x}=\left(\sum_{i=1}^{\infty}p^{-ix}\right)\left(\sum_{n \in G(S)}n^{-x}\right)$$
and solving the sum over $p^{-ix}$ we get
$$\sum_{n\in G(S\cup\{p\})}n^{-x}=\frac{1}{p^x-1}\left(\sum_{n \in G(S)}n^{-x}\right)$$
More generally, we get that, if we define for $S\subseteq \mathbb{P}$ the function $f(S)$ as the term
$$f(S)=\left(\prod_{p\in S}\frac{1}{p-1}\right)\cdot\left(\sum_{n\in G(S)}n^{-x}\right)$$
then
$$f(S\cup \{p\})=\frac{1}{(p^x-1)(p-1)}f(S)$$
and thus, by induction
$$f(S)=\prod_{p\in S}\frac{1}{(p^x-1)(p-1)}$$
which, if we let $N$ and $\varepsilon>0$ be such that for any $n>N$ it holds that $(n^x-1)(n-1)>n^{1+\varepsilon}$ and $S$ be a set with no elements less than $N$, then
$$f(S)\leq \prod_{p\in S}\frac{1}{p^{1+\varepsilon}}$$.
From here, we just want to find the desired sum, which is
$$\sum_{\substack{S\subseteq\mathbb{P}\\S\text{ is finite}}}f(S).$$
but, since $f(A\cup B)=f(A)f(B)$ for disjoint $A,B\subseteq \mathbb{P}$ as a direct result of the formula for $f$, we can split this to
$$\left(\sum_{S\subseteq(\mathbb{P}\cap[1,N])}f(S)\right)\left(\sum_{\substack{S\subseteq(\mathbb{P}\cap(N,\infty)) \\S\text{ is finite}}}f(S)\right)$$
However, since the right factor is strictly less than the product of the $S$ raised to the power of $-1-\varepsilon$ and no distinct $S$ have the same sum, the right factor cannot exceed $\sum_{n=1}^{\infty}n^{-1-\varepsilon}$, which is finite. The left factor is a sum of finitely many terms and is hence finite. Thus, the entire series converges.
