TITLE: An identity involving the Möbius function
QUESTION [7 upvotes]: $$\sum_{n=1}^{\infty}\frac{1}{n^s}\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=1$$
  for $s>1$.

How do I prove this identity?

REPLY [4 votes]: Let $D_f(s)$ denote the Dirichlet generating series:
$$
D_f(s) = \sum_{n =1}^{\infty} \frac{f(n)}{n^s}.
$$
When the series convergences absolutely, we can write $D_f(s)$ as an Euler product:
$$
D_f(s) = \prod_p \left(1+f(p)p^{−s} +f(p^2)p^{−2s} \dots \right)
$$
where the product ranges over all the primes $p$.  Looking at the Euler product for $\zeta(s)$ (which corresponds to $f(n) = 1$) we see that
$$
\zeta(s) = \prod_p \left(1+p^{-(s-1)} +p^{-2(s-1)}+ \dots \right) = \prod_p \left(1 - p^{-s} \right)^{-1}.
$$
Hence,
$$\begin{align}
\frac{1}{\zeta(s)} = \prod_{p} \left(1 - p^{-s} \right) = \sum_{n=1}^{\infty} \frac{g(n)}{n^s},
\end{align}$$
where $g$ is the multiplicative function such that $g(p) = -1$, and $g(p^n) = 0$ when $n \geq 2$.  Note that this is precisely the definition of $\mu$ (and this is probably how the formula for $\mu$ was originally discovered).  Thus, we have shown that:
$$ \sum_{n=1}^{\infty} \frac{1}{n^s} \cdot \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} = \zeta(s) \cdot \frac{1}{\zeta(s)} = 1$$
which is valid when $\zeta(s)$ converges absolutely (so this is valid for $s = \sigma + it \in \mathbb{C}$ such that $\sigma > 0$).