Consider the following problem:
Show that for $s>1$: $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}.$$
($\mu$ denotes the Mobius function)
My approach:
One may first note that the series converges for $s>1$:
$$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$ converges for $s_0=1$ thus it converges for $Re(s)>Re(s_0) $.
Now since $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$ we can write $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\frac{1}{\sum_{n=1}^{\infty}\frac{1}{n^s}}$$ thus we get
$$\tag{1}\bigl( \sum_{n=1}^{\infty}\frac{1}{n^s} \bigr) \bigl( \sum_{n=1}^{\infty}\frac{\mu(n)}{n^s} \bigr)=1$$
Let $\epsilon(n):=1 \forall n \in \mathbb{N}$ and $\eta(n):=1$ for $n=1$ and $0$ for $n \geq 2$.
For the Dirichlet Convolution $\ast$ and two arithmetic functions $a,b$ we know the following formula:
$$\bigl( \sum_{n=1}^{\infty}\frac{a(n)}{n^s} \bigr) \bigl( \sum_{n=1}^{\infty}\frac{b(n)}{n^s} \bigr)= \sum_{n=1}^{\infty}\frac{(a \ast b)(n)}{n^s}$$
Note that we can rewrite $(1)$ as $$\bigl( \sum_{n=1}^{\infty}\frac{\epsilon(n)}{n^s} \bigr) \bigl( \sum_{n=1}^{\infty}\frac{\mu(n)}{n^s} \bigr)=\sum_{n=1}^{\infty}\frac{\eta(n)}{n^s}.$$
Since $\epsilon \ast \mu = \eta$ we are already done.
My first question is, if this solution is correct.
I am interested in this kind of problem. Doing this specific problem, it was not that difficult since the only thing I needed to do is to write $\zeta(s)$ as a series and use the property of the Dirichlet-Convolution. Then the only thing left is to show the convolution identity. But what if only the left hand side is given and one has to find the right hand side/representation by $\zeta$?
For example, let $\tau(n):=\sum_{d |n} 1$, $\Lambda$ denote the von Mangoldt function (https://en.wikipedia.org/wiki/Von_Mangoldt_function) how would one solve an exercise of the form:
Represent the series through the $\zeta$ function for $$\sum_{n=1}^{\infty} \frac{\Lambda(n)}{n^s} ,\sum_{n=1}^{\infty} \frac{\tau(n)}{n^s} ,\sum_{n=1}^{\infty} \frac{\mu^2(n)}{n^s}$$ and so on...
How would one approach this kind of problem?
