It is mentioned in multiple occasions [here](https://terrytao.wordpress.com/tag/mobius-function/) that the bound
$$
\mathop{\sum\_{n=1}^{N}}\_{n\equiv a\mod l} \mu(n) = o(N)
$$
is equivalent to the prime number theorem in arithmetic progressions. But I am not able to find a proof of this equivalence (or the proof of the above bound) except for the case when $l=1$ which is the classical prime number theorem. If anyone can point me to a reference, it will be much appreciated.


Also, there is a result in [Edwards' book](https://www.google.co.in/books/edition/_/-66eHCf2Y8kC?hl=en&gbpv=0) where he proves the result
$$
\left|\sum\_{n=1}^{N} \frac{\mu(n)}{n}\right| \leq \frac{K}{\log(N)}.
$$
Is there an analogue of this bound for arithmetic progressions as well?


There are some discussions [here](https://mathoverflow.net/questions/356251/analytic-equivalents-for-primes-in-arithmetic-progressions), but they don't consider the Möbius function



Averages of Möbius function in arithmetic progressions
When $(a,m) = 1$, let $\pi(x;a \bmod m)$ be the number of primes $p\leq x$ such that $p \equiv a \bmod m$.
The prime number theorem for arithmetic progressions
mod $m$ says for all $a \in (\mathbf Z/m\mathbf Z)^\times$ that $\pi(x;a \bmod m) \sim (1/\varphi(m))x/\log x$.


Harold Shapiro, in the paper *Some assertions equivalent to the prime number theorem for arithmetic progressions*, Comm. Pure Appl. Math. 2 (1949), 293-308, showed
that theorem for an integer $m \geq 1$ is equivalent to each of the following conditions for the same $m$:
$$
\sum\_{\substack{n \leq x \\ n \equiv a \bmod m}} \mu(n) = o(x) 
$$
for all $a$ with $(a,m) = 1$ and
$$
\sum\_{\substack{n \geq 1 \\ n \equiv a \bmod m}} \frac{\mu(n)}{n} \ \ {\rm converges}
$$
for all $a$ with $(a,m) = 1$. This should address both of your questions.


Concerning the second equivalent condition above, note that the prime number theorem is in many places expressed as being equivalent to the calculation $\sum \mu(n)/n = 0$, but it is also equivalent just to the convergence of $\sum \mu(n)/n$ because it is easy to show that this series must be $0$ if it converges thanks to Abel's theorem for Dirichlet series: convergence of $\sum \mu(n)/n$ implies this series must equal
$\lim\_{s \to 1^+} \sum \mu(n)/n^s = \lim\_{s \to 1^+} 1/\zeta(s) = 0$.


I explained in my answer [here](https://mathoverflow.net/questions/356251/analytic-equivalents-for-primes-in-arithmetic-progressions) how the prime number theorem in arithmetic progressions mod $m$ is equivalent to nonvanishing of $L(s,\chi)$ on the line ${\rm Re}(s) = 1$
for all Dirichlet characters $\chi \bmod m$,
With this information, let's see how to derive the first Moebius analogue above. When $(a,m) = 1$,
$$
\sum\_{\substack{n \leq x \\ n \equiv a \bmod m}} \mu(n) = \sum\_{n \leq x} \frac{1}{\varphi(m)}\left(\sum\_{\chi \bmod m} \overline{\chi}(a)\chi(n)\right)\mu(n), 
$$
which is
$$
\frac{1}{\varphi(m)}\sum\_{\chi \bmod m} \overline{\chi}(a)\left(\sum\_{n \leq x} \chi(n)\mu(n)\right).
$$
To show this is $o(x)$, we show each inner sum is $o(x)$.


The inner sum at $\chi$ is the partial sum of the coefficients of the Dirichlet series $\sum \chi(n)\mu(n)/n^s$, which is $1/L(s,\chi)$ for ${\rm Re}(s) > 1$.
The coefficients of this Dirichlet series are bounded in absolute value (by $1$) and $1/L(s,\chi)$ has an analytic continuation to the line ${\rm Re}(s) = 1$ (this is where we use the nonvanishing of all $L(s,\chi)$ on that line, including the pole at $s = 1$ when $\chi$ is the trivial character, so the reciprocal there is analytic at $s=1$ with a zero), so by Newman's Tauberian theorem
$$
\frac{1}{x}\sum\_{n \leq x} \chi(n)\mu(n) \to {\rm Res}\_{s=1} \frac{1}{L(s,\chi)} = 0.
$$
The convergence of $\sum\_{n \equiv a \bmod m} \mu(n)/n$ when $(a,m) = 1$ is proved by a similar approach: reduce the task to showing for all Dirichlet characters $\chi \bmod m$ that $\sum\_{n \leq x, n \equiv a \bmod m} \chi(n)\mu(n)/n$ converges as $x \to \infty$.


