 Solution

The Möbius function has $\mu (1)=1$, vanishes on integers divisible by a square of a prime, and equals $(-1{)}^r$ on a product of $r$ distinct primes. Factoring the divisor sum prime by prime givesThis is the Möbius divisor-sum identity.

For $T/2\le t\le T$ and $0<\sigma \le 1$, the Hardy-Littlewood approximation to the Riemann zeta function at cutoff $T$ givesIndeed $T\ge |t|/\pi$, and the omitted integral term has size at most $T^{1-\sigma }/|s-1|\ll T^{-\sigma }$. Multiply by ${\sum }_{m\le M}\mu (m)m^{-s}$. Its absolute value is at mostwhere the elementary inequality follows from $m^{1-\sigma }\le M^{1-\sigma }$. Reindexing the finite double sum yields coefficients $a_n={\sum }_{m\mid n,m\le M,n/m\le T}\mu (m)$, with no terms for $n>MT$. For $n\le \min (M,T)$ all divisors meet both restrictions, so the Möbius divisor-sum identity gives $a_1=1$ and $a_n=0$ for $1<n\le \min (M,T)$. Hence the truncated Möbius inverse identity for the Riemann zeta function isThe displayed error is uniform in $0<\sigma \le 1$ and the stated height interval.