using Dirichlet transforms to show infinity of primes I am trying to understand the following lemma and its proof. The lemma and proof are on page 2 of this pdf: http://www.ias.ac.in/resonance/March2012/p284-290.pdf
Basically, I want to know their reasoning of switching sums? Why go over $n$ first, but then go over $r$ after and what is "r" anyways. 
Also I am confused about they can apply the formula for geometric series. 
And my final question is how does this (combined with the Möbius formula) imply infinitude of primes? 
 A: Ah, probably a typo.. I think, this should be correct:
$$F(z)=\sum_{r\ge 1} \hat f(r)\, z^r=\sum_{r\ge 1}\sum_{d|r}f(d)\,z^r=\sum_{d\ge 1}\sum_{n\ge 1}f(d)z^{nd}=\sum_df(d)\cdot \frac{z^d}{1-z^d}\,.$$
Update: In the  finalized version  (which you mentioned only in a comment), these formulas are all correct, and the conclusion for having infinitely many primes follows:
Consider the convolution unit: $\hat u(1)=1$ and $\hat u(n)=0$ if $n>1$, with that, by the inversion formula, we have
$$u(n)=\sum_{d|n} \mu(n/d)\,\hat u(d)=\mu(n)\,.$$
Now the support of $\hat u$ is $\{1\}$, finite, so, by the lemma the support of $u=\mu$ must be infinite. However, if we had only finitely many primes, there would be only finitely many square-free numbers (all the products of distinct primes), so $\mu$ would be of finite support, a contradiction.
A: Yes, it seems there are typos in the Lemma.  The reasoning for switching the sums is that the series for $F(z)$ is absolutely convergent.  The idea is you want to sum over $rd = n$, but the last equation line in the Lemma is not written correctly.
I do not see how the infinitude of the primes follows from the fact that $\mu$ has infinite support.
