Prove $ \left\vert \sum_{n=1}^N \frac{\mu(n)}{n} \right\vert \leqslant 1,$ where $\mu(n)$ is the Mobius function. Given a positive integer $N$, show that 
$$ \left\vert \sum_{n=1}^N \frac{\mu(n)}{n} \right\vert  \leqslant 1,$$
where $\mu(n)$ is the Mobius function.
How do I approach this question? I guess a particular property of the Mobius function might be involved in this question, but I am not able to crack it.
 A: This can be done without PNT. 
From my answer in this one: Prove $\sum_{d \leq x} \mu(d)\left\lfloor \frac xd \right\rfloor = 1 $
We have for $x\geq 1$, 
$$
\sum_{d\leq x} \mu(d) \left\lfloor \frac xd \right \rfloor =1
$$
Write the LHS as 
$$
\sum_{d\leq x} \mu(d) \left( \frac xd -\left\{ \frac xd \right\} \right) = 1
$$
where $\{ x\}$ is the fractional part of $x$. 
Then we have
$$
\sum_{d\leq x } \mu(d) \frac xd = 1+\sum_{d\leq x} \mu(d)\left\{ \frac xd \right\}.
$$
If $x$ is a positive integer, then the RHS is bounded by
$$
\left | 1+\sum_{d\leq x} \mu(d)\left\{ \frac xd \right\}\right|\leq 1+ x-1=x. 
$$
Note that $\left\{\frac xx\right\} = 0$. 
Therefore if $x$ is a positive integer, 
$$
\left| \sum_{d\leq x} \mu(d)\frac xd \right|\leq x.
$$
This immediately gives for $x$ positive integer, 
$$
\left| \sum_{d\leq x} \frac{\mu(d)}d\right|\leq 1
$$
A: For any $s>1$ we have $\sum_{n\geq 1}\frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$ and since the $\zeta$ function has a simple pole at $s=1$ we also have $\sum_{n\geq 1}\frac{\mu(n)}{n}=0$ (this is essentially equivalent to the PNT). We may also consider that for any $s>1$
$$ \sum_{n=1}^{N}\frac{\mu(n)}{n^s}\sum_{n\geq 1}\frac{1}{n^s}=\sum_{n\geq 1}\frac{1}{n^s}\sum_{\substack{d\mid n\\ d\leq N}}\mu(d)=1-\sum_{n>N}\frac{1}{n^s}\sum_{\substack{d\mid n\\d> N}}\mu(d) $$
by Moebius inversion formula, where $d(n)\ll n^\varepsilon$ for any $\varepsilon>0$. In particular the crude inequality
$\left|\sum_{\substack{d\mid n\\ d>N}}\mu(d)\right|\leq d(n)$ already ensures
$$\left|\sum_{n=1}^{N}\frac{\mu(n)}{n^s}\right|\ll \frac{1}{\zeta(s)}\left(1+\frac{N^{1+\varepsilon-s}}{s-1}\right)=N^{\varepsilon}\quad\text{as }s\to 1^+ $$
and one just needs to exploit a weak cancellation in $\sum_{\substack{d\mid n\\ d>N}}\mu(d)$ to be sure that the partial sums $\sum_{n=1}^{N}\frac{\mu(n)}{n}$ are bounded. "Bounded by $1$" can be checked by hand once $\lim_{N\to +\infty}\sum_{n=1}^{N}\frac{\mu(n)}{n}=0$ (the PNT) is granted.
