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.


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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.