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This is a problem from Intro to Arithmetic functions by Paul Mccarthy.

For all $n,\sum_{1\leq j\leq n}\mu(j)[\frac{n}{j}]=1$.

Since $\mu$ is multiplicative, I know that only products of distinct primes count non-trivial terms. However I do not know how to handle $[\frac{n}{\prod_{i}p_i}']$ where $p_i$ are distinct and $\prod_ip_i\leq n$. $[x]$ is the greatest integer $n$ s.t. $n\leq x$.

How should I progress? The book has not talked about riemann zeta function yet. So I guess I am not allowed to convolve $\zeta$ with $\mu$.

user45765
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  • @DietrichBurde This is indeed not a divisor sum. I can break the sum into two parts. One is a sum of divisors' contribution and the other one is not. In that way, I do not know how to deal with integer part. Any hint will be helpful. – user45765 Oct 28 '17 at 19:04
  • $$\sum_{j=1}^n\mu(j)\Bigl\lfloor\frac{n}{j}\Bigr\rfloor=\sum_{j=1}^n\mu(j)\sum_{j\mid i\le n}1=\sum_{i=1}^n\sum_{j\mid i}\mu(j)$$ – Mercury Oct 28 '17 at 19:41

1 Answers1

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For arithmetic functions we have the Dirichlet product $$ \sum\limits_{d|n} \mu(d)\nu(n/d)=1, $$ where $\nu(n)$ denotes the number of positive divisors of $n$. This is proved here:

Sum of Positive Divisors: $\sum_{d|n} \mu(n/d)\nu(d)=1$ and $\sum_{d|n} \mu(n/d)\sigma(d)=n$.

We also have (see the comment above, and here) $$ \sum_{j\mid i}\mu(j)=\left\lfloor \frac{1}{i}\right\rfloor , $$ so that $$ \sum_{j=1}^n\mu(j)\left\lfloor \frac{n}{j}\right\rfloor =\sum_{i=1}^n\sum_{j\mid i}\mu(j)=\sum_{i=1}^n \left\lfloor \frac{1}{i}\right\rfloor =1. $$

Dietrich Burde
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