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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.

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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, $$ \left| \sum_{d\leq x} \frac{\mu(d)}d\right|\leq 1 $$

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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.

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  • $\begingroup$ thanks for the solution but I haven't learnt anything about the zeta function, any method not involving the zeta function $\endgroup$ – saisanjeev Feb 20 '18 at 5:46
  • $\begingroup$ @saisanjeev: not to involve the zeta function is just a way for complicating things, like in the elementary proof of the PNT. On the other hand, what is the original reason for studying $\sum \frac{\mu(n)}{n}$ if not in relation with the zeta function? $\endgroup$ – Jack D'Aurizio Feb 20 '18 at 11:03
  • $\begingroup$ I just came across this question in Berton's Book on Number Theory, perhaps will get back to this question as soon as possible after going through the zeta function once $\endgroup$ – saisanjeev Feb 20 '18 at 14:45

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