Dirichlet convolution for dummies Can someone explain me meaning and usefulness of Dirichlet convolution?
I know the concepts of summation function over divisors of number and its Moebius inversion. But how does it relate to Dirichlet convolution? Concrete examples would be helpful.
Thanks!
 A: Let us suppose you have two different absolutely convergent Dirichlet series
$$ A(s) = \sum_{n \geq 1} \frac{a(n)}{n^s} \qquad \text{ and } \qquad B(s) = \sum_{m \geq 1} \frac{b(m)}{m^s},$$
and you want to multiply them together. Let's write out the first few terms of each:
$$A(s) = a(1) + \frac{a(2)}{2^s} + \frac{a(3)}{3^s} + \ldots$$
$$B(s) = b(1) + \frac{b(2)}{2^s} + \frac{b(3)}{3^s} + \ldots$$
Then 
$$\begin{align}
A(s)B(s) &= \left(a(1) + \frac{a(2)}{2^s} + \frac{a(3)}{3^s} + \ldots\right)\left(b(1) + \frac{b(2)}{2^s} + \frac{b(3)}{3^s} + \ldots\right) \\
&=a(1)b(1) + \frac{a(1)b(2) + a(2)b(1)}{2^s} + \frac{a(1)b(3) + a(3)b(1)}{3^s} + \\
&\quad + \frac{a(1)b(4) + a(2)b(2) + a(4)b(1)}{4^s} + \ldots
\end{align}$$
If you think about it, the coefficients of the product are pretty easy to write down. The $k$th coefficient is formed by those products of denominators that multiply to $n$, and any such product of denominators that multply to $k$ contribute the the $k$th coefficient. In particular, the $k$th coefficient of $A(s)B(s)$ is given by
$$ \dfrac{\displaystyle\sum_{mn = k}a(n)b(m)}{k^s}.$$
And the numerator is a Dirichlet convolution $a \star b(k)$. This is why Dirichlet convolutions are natural objects to think about.
When dealing with multiplicative functions, when $A(s)$ and $B(s)$ have Euler products, then it is extremely easy and sometimes useful to interplay Euler product manipulations, Dirichlet convolution manipulations, and products of Dirichlet series. 
For example, we know that $\displaystyle \zeta(s) = \sum_{n \geq 1} \frac{1}{n^s} = \prod_p \left(1 - \frac{1}{p^s}\right)^{-1},$ so we could write $\displaystyle \frac{1}{\zeta(s)} = \prod_p \left(1 - \frac{1}{p^s}\right).$ Whenever this converges absolutely (for $\Re s > 1$), this Euler product can be multiplied out into a Dirichlet series which we denote by $\displaystyle\sum_{m \geq 1} \frac{\mu(m)}{m^s}$. From the Euler product, we see that $\mu(m)$ is not supported on prime powers, but only on primes and products of primes. In fact, $\mu(p) = -1$, $\mu(pq) = 1, \mu(pqr) = -1, \ldots$.
If I call $M(s) = \displaystyle \sum_{m \geq 1} \frac{\mu(m)}{m^s}$, then we have that $\displaystyle \frac{1}{\zeta(s)} = M(s)$, or equivalently $\displaystyle \zeta(s)M(s) = 1$. 
In terms of Dirichlet convolutions, if we denote by $1(n)$ the function that takes everything to $1$ and $\varepsilon(n)$ the function that is $1$ on $1$, and $0$ everywhere else, then this last product of series is the same thing as 
$$ 1\star \mu(n) = \varepsilon(n).$$
This is what allows Mobius inversion to work. In particular, if $f$ and $g$ are functions such that $f(n) = \displaystyle \sum_{d \mid n} g(d)$, or equivalently $f(n) = 1\star g (n)$, then convolving with $\mu$ gives that 
$$f \star \mu(n) = g(n),$$
which is used all the time in analytic number theory. This is just an example of the interplay between Euler products, Dirichlet series, and Dirichlet convolution. If you pick up an introductory book on analytic number theory, such as Apostol's, and browse through it, you'll see that these form the bread and butter of the techniques available - ultimately leading to the methods used to bound, estimate, and understand most of the arithmetic functions of analytic number theory.
