Asymptotics of $\sum_1^x \mu(n)/n$

Question

It is known that $\Bigl|{\sum_1^x \mu(n)/n}\Bigr|\leq 726/\log^2 x$ for $x>1$, reference here.

A) Is there a result on the l0wer bound $f$ of $\sum_1^x \mu(n)/n$ of the type ${\lim\sup}_{x \to \infty}\Bigl|{\sum_1^x \mu(n)/n}\Bigr|\geq O(f)$?

B) Could you please locate the error in the following:

a) $\zeta(s)=\sum_1^\infty 1/n^s$
-- this is true for $Re(s)>1$;
b) $\zeta(1)=\lim_{s \to 1^+}\zeta(s)=\lim_{s \to 1^+}\sum_1^\infty 1/n^s=\sum_1^\infty 1/n$
-- the limit does not exist, but this expression nonetheless describes the pole of zeta at $1$;
c) $\zeta(1)=\lim_{x \to \infty}\sum_1^x 1/n$
-- this seems to follow naturally from b);
d) $\sum_1^x 1/n = H(x)\sim \log x$
-- this is just the definition of a harmonic number $H(x)$ and its well known asymptotics;
e) $\zeta(1)\sim\log x$, or, more precisely, $\zeta(1)=\lim_{x\to\infty}\log x$
-- this simply follows from c) and d);
f) $\sum_1^\infty \mu(n)/n^s=1/\zeta(s)$
-- this is true at least for $Re(s)\geq 1$;
g) $\sum_1^\infty \mu(n)/n=1/\zeta(1)$
-- this doesn't seem problematic, since it follows from f);
h) $\lim_{x\to\infty}\sum_1^x \mu(n)/n=\lim_{x\to\infty}\frac{1}{\sum_1^x 1/n}$
-- this follows from g) and c);
i) $\sum_1^x \mu(n)/n\sim 1/\log x$
-- this follows from h) and d).

There is an error in here in B), because the result i) is in a sharp contrast with the known better result $\Bigl|{\sum_1^x \mu(n)/n}\Bigr|\leq 726/\log^2 x$, $x>1$. Can someone point a finger at the exact location of the error?

Answer 1

By definition of $\mu(n)$ : $$(\sum_{m=1}^\infty \mu(m)m^{-s})(\sum_{k=1}^\infty k^{-s}) = \sum_{m \ge 1, k \ge 1} (km)^{-s} \mu(m) = \sum_{n=1}^\infty n^{-s} \sum_{d | n} \mu(d) = 1$$

This is true as formal series, and as convergent series only for $Re(s) > 1$.

Now you can write that

$$(\sum_{m=1}^N \mu(m)m^{-s})(\sum_{k=1}^N k^{-s}) = \sum_{1 \le m \le N, 1 \le k \le N} (km)^{-s} \mu(m)$$ $$ = \sum_{n=1}^{N^2} n^{-s} \sum_{d | n, d \le N, n/d \le N} \mu(d) = 1 + \sum_{n = N+1}^{N^2}n^{-s}\sum_{d | n, d \le N,n/d \le N} \mu(d) \ne 1$$

Hence $\sum_{n = 1}^N n^{-s} \sim f(N)$ doesn't imply that $\sum_{n = 1}^N \mu(n) n^{-s} \sim \frac{1}{f(N)}$