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This question already has an answer here:

I'm currently sitting with the following number theory problem:

Prove that for all $X \geq 1$, $$\sum_{1 \leq n \leq X} \mu(n) \left[\frac{X}{n}\right] = 1.$$

A few ideas I have tried: Using that $[x]=x - \{x\}$ which I couldn't make out to bring anything. Switching around on the summation variables, like $$\sum_{d \leq N} \mu(d) f\left( \frac{n}{d} \right) = \sum_{d \leq N} \mu \left( \frac{d}{n} \right) f(d)$$ but that didn't seem to bring anything.

I tried writing out the first few terms, which give $$\mu(1) [X] - \left[\frac{X}{2}\right] - \left[\frac{X}{3}\right] - \left[\frac{X}{5}\right] + \left[\frac{X}{6}\right] - \left[\frac{X}{7}\right] - \left[\frac{X}{8}\right]$$ but I can't seem to find a system where it simply reduces to one!

What do you think?

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marked as duplicate by user91500, Tom-Tom, Venus, Micah, Erick Wong Jun 29 '15 at 16:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Claim: if $f$ is any function, then

$$\sum_{n=1}^X f(n)\left\lfloor\frac{X}{n}\right\rfloor = \sum_{n=1}^X \sum_{d\mid n} f(d).$$

Proof: By rearrangement. How many times does $f(d)$ appear in the sum on the right-hand side?

Alternatively, proceed by induction on $X$ using the fact that $$\left\lfloor\frac{X+1}{n}\right\rfloor - \left\lfloor\frac{X}{n}\right\rfloor$$ equals $1$ if $n\mid (X+1)$ and $0$ otherwise.

Corollary: for all $X\geq 1$, $$\sum_{n=1}^X \mu(n)\left\lfloor\frac{X}{n}\right\rfloor = 1.$$

Indeed, $\sum_{d\mid n} \mu(d)$ is $1$ for $n=1$ and $0$ for $n>1$.

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Use this theorem.

Theorem: Let $f$ and $F$ be number-theoretic functions such that $$F(n)=\sum_{d|n}f(d),$$ Then for any positive integer $X$, $$\sum_{n=1}^X F(n)=\sum_{n=1}^X f(n)\left[\frac{X}{n}\right].$$

$$\sum_{1 \leq n \leq X} \mu(n) \left[\frac{X}{n}\right] =\sum_{n=1}^X F(n), F(n)=\sum_{d|n}\mu(d)$$ $$\sum_{n=1}^X\sum_{d|n}\mu(d)=\sum_{d|1}\mu(d)+\sum_{n=2}^X\sum_{d|n}\mu(d)=1+0=1.$$

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