Euler product of a Dirichlet function

Quote:

Originally Posted by Deadstar

I'm so close to this answer but just can't quite get it 100%...

Show that if $f(n)$ is a strongly multiplicative function then the Euler product of its Dirichlet function $D_f (s)$ is of the form

$\prod_p \left ( 1 - \frac{f(p)}{p^s} \right )^{-1}$

What I know...

Since $f(n)$ is strongly multiplicative... It can be written as,

$\prod_p f(p)$ where p is a prime number. (1)

The Dirichlet series I'm after is of the form,

$\sum_{n=1}^{\infty} f(n) n^{-s}$ .

Now here's where I kinda get lost. I'll write down exactly what I've been writing down as my answer.

$\sum_{n=1}^{\infty} f(n) n^{-s} = \frac{1}{1 - \frac{f(n)}{n^s}}$ by the geometric series.

So by (1) we can write this as...

$\prod_p \left (1-\frac{f(p)}{p^s} \right )^{-1}$

Is this right? Something about it feels wrong, like I'm not being very rigorous or that I've missed a step...

It looks just fine to me: since the function is completely (or totally or, as you say, strong) multiplicative, it's easy to see that $\sum^\infty_{n=1}f(x)=\prod^{}_{p,\,a\,\,prime}\le ft(1-f(p)\right)^{-1}$ , so its Dirichlet series $\sum^\infty_{n=1}\frac{f(n)}{n^s}$ fulfills what you wrote.

Tonio