Find the closed form of the the sums:

Question

I am trying to find the closed form of the following sums:

a) $$\sum_{r|d}\mu(r)\tau\left(\frac{n}{r}\right)$$ b) $$\sum_{d|n}\mu(d)\sigma\left(\frac{n}{d}\right)$$ c) $$\sum_{r|n}\mu(r)\tau(r)$$ d) $$\sum_{d|n}\mu(d)\sigma(d)$$

For a), I suspect that we have $1^k$ ($k$ is the quantity of the divisors of $n$).

For b), we just have $n$.

I am hoping to prove this, but I got stuck.

Answer 1

You want to reverse the Möbius inversion formula to evaluate the first two of these. It says that if $f$ and $g$ are arithmetic functions such that $$g(n)=\sum_{d|n}f(d)$$ then $$f(n)=\sum_{d|n}\mu(d)g(n/d)$$

If you can guess the correct answer, this formula will make proving correctness easy.