How does one show the following estimate:
$\displaystyle \bigg\lvert 1+\sum_{n \leq x} \mu(n) \Big\{\frac{x}{n}\Big\} \bigg\rvert \leq x$, where $x\geq1$ ?
My attempt was to use the triangle inequality so that
$\displaystyle \bigg\lvert 1+\sum_{n \leq x} \mu(n) \Big\{\frac{x}{n}\Big\} \bigg\rvert \leq 1 + \sum_{n \leq x} \Big\{\frac{x}{n}\Big\} = 1 + \{x\} + \sum_{2\leq n \leq x-1} \Big\{\frac{x}{n}\Big\}$,
and to try and bound the re-indexed sum by $[x]-1$ (here, $[x]$ denotes the floor of x), which will then give $1+\{x\}+[x]-1=\{x\}+[x]=x$. Dividing by x would then yield the desired result, but I'm having difficulty in finding why this bound makes sense. Am I missing something obvious here? Perhaps there's an easier way to derive this? I also suspect that fiddling around with the triangle inequality is unnecessary, and this could be shown directly without much effort, but I'm just not seeing it. Any explanations would be appreciated!
Best Answer
First, I want to point out that $$\sum_{n\leq x}\left\{\frac nx\right\}\neq\{x\}+\sum_{2\leq n\leq x-1}\left\{\frac nx\right\}.$$ The problem is that $x$ need not be an integer (in particular, you may not have a case $n=x$, so $x-1$ might miss an integer). Now, the fractional part of a number is always bounded above by $1$, so you in fact have $$\sum_{2\leq n\leq x}\left\{\frac xn\right\}\leq\sum_{2\leq n\leq x}1=\lfloor x\rfloor-1.$$ Combining the steps as you have, you end up with $$\left|1+\sum_{n\leq x}\mu(n)\left\{\frac xn\right\}\right|\leq 1+\{x\}+\lfloor x\rfloor-1=x.$$