So my question is:
Use the Dirichlet series to show that
$\sum_{k|n}\mu(k)d(\frac{n}{k})$ = 1
for all natural numbers n where d(.) is the divisor function.
I've just started learning about the Dirichlet convolution and I gather that
$\sum_{k|n}\mu(k)d(\frac{n}{k})$ = $\mu$ $\star$ d (n) = $\sum_{k|n}\mu(\frac{n}{k})d(k)$ but I don't know how to use that to prove that the LHS = 1
We can write $$ \sum_{k|n}\mu(k)d\left(\frac{n}{k}\right) = \mu *d $$ where $*$ is Dirichlet convolution. The Dirichlet series of $\mu(n)$ is $$ \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} $$ The Dirichlet series of $d(n)$ is $$ \sum_{n=1}^\infty \frac{d(n)}{n^s} = \zeta(s)^2 $$ the convolution of the two is given by the product of their series $$ \mu * d = \frac{1}{\zeta(s)}\cdot \zeta(s)^2 =\zeta(s) $$ and $\zeta(s)$ is the Dirichlet series for $1$ $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$ thus $$ \mu *d = \sum_{k|n}\mu(k)d\left(\frac{n}{k}\right) = 1 $$