 Solution

The möbius function is $mu(1)=1$, $mu(n)=0$ if a prime square divides $n$, and $mu(n)=(-1{)}^r$ when $n$ is a product of $r$ distinct primes. The Riemann zeta function is $zeta(s)={\sum }_{n\ge 1}{n}^{-s}$ for $\Re s>1$.

Both sides of the proposed identity are multiplicative. At $n=p^a$, the right side is $1$ for $a=0,1$ and $1+\mu (p)=0$ for $a\ge 2$, exactly $mu(p^a{)}^2$. Absolute convergence permits rearrangement, so