Proving associativity of product of two formal sums $\sum_{n = 1}^{\infty} \frac{a_n}{n^x}$ Let $R$ be the set of all formal sums $\sum_{n = 1}^{\infty} \frac{a_n}{n^x}$ where $a_n \in \Bbb{Q}$, where two sums $a, b$ are equal iff $a_i = b_i \ \forall i$.  It is indeed a ring with addition defined by $a + b = \sum_{n=1}^{\infty} \frac{a_n + b_n}{n^x}$, and multiplication defined by $a\cdot b = \sum_{n=1}^{\infty} \dfrac{\sum_{d \mid n} a_d b_{n/d}}{n^x}$.  Proof: Clearly an additive abelian group.  It has a a multiplicative identity given by $e_n = 0 \ \forall n \gt 1$ and $e_1 = 1$ so that the associated sum is $\dfrac{1}{1^x}$.  Multiplication is associative:
$$(a\cdot (b \cdot c))_n = \sum_{d \mid n} a_d(b\cdot c)_{n/d} = \sum_{d \mid n} a_d\sum_{d'\mid n/d} b_{d'} c_{n/d'}.$$  Here is where I'm stuck.
 A: Associativity becomes clearer when you write the Dirichlet convolution as $$\sum_{d \mid n} a_d b_{n/d} = \sum_{d \cdot e=n} a_{d} b_{e}$$
Then we write
$$\sum_{d \cdot g = n} a_d \sum_{e \cdot f=g} b_{e} c_{f} = \sum_{d\cdot e \cdot f = n}a_d b_{e} c_{f}$$
Associativity follows, since you can write $$\sum_{d \cdot g = n} \left(\sum_{e \cdot f = g} a_e b_f\right) c_d=\sum_{d\cdot e \cdot f = n}a_e b_{f} c_{d}.$$
The two sums on the right differ only by labeling. Thus the operation is associative.
A: Your sum is over all $d,d'$ such that $d$ divides $n$ and $d'$ divides $n/d$. 
Consider the fact that given such $d$ and $d'$, this uniquely determines a number $d''$ such that $d \cdot d' \cdot d'' = n$ (i.e. $d''=n/(d\cdot d')$). 
Likewise, given divisors of $n$: $d,d',d''$ such that $d \cdot d' \cdot d'' = n$, we have that $d$ divides $n$, $d'$ divides $n/d$ (since $d' \cdot (d \cdot d'') = n$), and $d'' = n/d'$
This means that...
$$(a\cdot (b \cdot c))_n = \sum_{d | n} a_{d}(b\cdot c)_{n/d} = \sum_{d | n} a_{d}\sum_{d'| n/d} b_{d'} c_{n/d'} = \sum_{d | n} \sum_{d'| n/d} a_{d}b_{d'} c_{n/d'} = \sum_{(d,d',d'') \in S} a_d b_{d'} c_{d''}$$
where $S = \{ (d,d',d'') \in \mathbb{Z}_{>0}^3 \;|\; d \cdot d' \cdot d'' = n\}$.
It shouldn't be hard to show that $((a \cdot b) \cdot c)_n$ is equal to the same end result. :)
