Background & Question
I recently thought of a combinatoric method to get an interesting result:
$$ \sum_{r=n+1}^{n!} \sum_{\substack{k|r \\ k \leq n}} \mu \left({ {k}}\right) = n! O(\frac{1}{\sqrt n}) -1 $$
I was wondering how other methods compared to mine in calculating the quantity:
$$\sum_{\substack{k|r \\ k \leq n}} \mu \left({ {k}}\right)$$
and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?
Proof of my method
Consider the following sums: $$ 1+1 + 1+ 1+1 +\underbrace\dots_{n! \text{ times}} = n! $$ $$ 0 + 1+ 0+1 +0 +\underbrace\dots_{n! \text{ times}} = \frac{n!}{2} $$ $$ 0 + 0+ 1+0 +0 +\underbrace\dots_{n! \text{ times}} = \frac{n!}{3} $$ We proceed to do so $n$ times: $$ 0+ 0 +\underbrace\dots_{n \text{ times}} +1+\underbrace\dots_{n! \text{ times}} = \frac{n!}{n} $$
Multiplying the $r$th row with $a_r$ which is an arbitrary variable:
$$ a_1+ a_1 + a_1+ a_1+a_1 +\underbrace\dots_{n! \text{ times}} = n! a_1 $$ $$ 0 + a_2+ 0+a_2 +0 +\underbrace\dots_{n! \text{ times}} =n! \frac{a_2}{2} $$ $$ 0 + 0+ a_3+0 +0 +\underbrace\dots_{n! \text{ times}} = n! \frac{a_3}{3} $$ We proceed to do so $n$ times: $$ 0+ 0 +\underbrace\dots_{n-1 \text{ times}} +a_n+\underbrace\dots_{n! \text{ times}} = n!\frac{a_n}{n} $$
Adding all the above equations vertically (as it is a finite sum):
$$ \underbrace{b_1}_{a_1} + \underbrace{b_2}_{a_1+a_2}+ \underbrace{b_3}_{a_1+a_3} + \dots+ b_n +\sum_{r=n+1}^{n!} \tilde b_r = n!\sum_{r=1}^n \frac{a_r}{r}$$
Writing the above properly:
$$ \sum_{r=1}^n b_r + \sum_{r=n+1}^{n!} \tilde b_r = n! \sum_{r=1}^n \frac{a_r}{r} $$
where $b_r = \sum_{r|k} a_k$ and $\tilde b_r = \sum_{\substack{k|r \\ k\leq n}} a_k $. Using the mobius function we know $ a_{n}=\sum_{d\mid n}\mu \left({\frac {n}{d}}\right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:
$$ \sum_{r=1}^n b_r + \sum_{r=n+1}^{n!} \sum_{\substack{k|r \\ k\leq n}} \sum_{d\mid k}\mu \left({\frac {k}{d}}\right)b_{d} = n! \sum_{r=1}^n \frac{\sum _{d\mid r}\mu \left({\frac {r}{d}}\right)b_{d}}{r} $$
Applying $\frac{\partial}{\partial b_1} $ both sides:
$$ 1 + \sum_{r=n+1}^{n!} \sum_{\substack{k|r \\ k \leq n}} \mu \left({ {k}}\right) = n! \sum_{r=1}^n \frac{\mu \left({{r}}\right)}{r} $$
Using RH:
$$ 1 + \sum_{r=n+1}^{n!} \sum_{\substack{k|r \\ k \leq n}} \mu \left({ {k}}\right) = n! O(\frac{1}{\sqrt n}) $$