Show that $\sum_{n=1}^{\infty}\cfrac{1}{n^s}\cdot \sum_{n=1}^{\infty}\cfrac{\mu(n)}{n^s}=1$ Show that $$\sum_{n=1}^{\infty}\cfrac{1}{n^s}\cdot \sum_{n=1}^{\infty}\cfrac{\mu(n)}{n^s}=1$$
Where $\mu(n)$ is Möbius function: $\mu(n)=\begin{cases}
1~~~~~~~if ~~n=0\\
0  ~~~~~~~  if~ n~ is~ divisible ~by ~a ~square ~larger~ than ~1\\
(-1)^m~~~~~~~If~~n~has~m~prime~factor
\end{cases}$
 A: I don't know if you are allowed to do this for your proof, but you can use the zeta function.
\begin{equation}
\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}
\end{equation}
You can also use the Euler product of the zeta function to find a generating function for the zeta function using the mobius function where
\begin{equation}
\sum_{n = 1}^\infty \frac{\mu (n)}{n^s} = \frac{1}{\zeta(s)}
\end{equation}
You can find the proof for this relation here. From this you can see that
\begin{align}
\sum_{n = 1}^\infty \frac{1}{n^s} \cdot \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \zeta(s) \cdot \frac{1}{\zeta(s)} = 1
\end{align}
A: In comments r9m gave a good hint.
Let, $f,g:\mathbb{Z}_+ \rightarrow \mathbb{C}$ be two funtions.
Define, $f*g:\mathbb{Z}_+ \rightarrow \mathbb{C}$ as, $$f*g(n)=\sum \limits_{d|n} f(d)g\left(\frac{n}{d}\right)$$
If we write, $f=\sum \limits_{n=1}^\infty f(n)\cdot n$ (here $f(n)\cdot n$ doesn't mean multiplication of complex numbers) and $g=\sum \limits_{n=1}^\infty g(n)\cdot n$, we can see, 
$$\sum \limits_{n=1}^\infty f*g(n)\cdot n=\left(\sum \limits_{n=1}^\infty f(n)\cdot n\right)\cdot\left(\sum \limits_{n=1}^\infty g(n)\cdot n\right)$$
We can also use the notaion $f=\sum \limits_{n=1}^\infty f(n)\cdot c(n)$ and $g=\sum \limits_{n=1}^\infty g(n)\cdot c(n)$, where $c$ is any completely multiplicative function.
Here ‘$\cdot $’ can be viewed as multiplication of complex numbers if all the above series converges absolutely.
Here in this question $c(n)=\dfrac{1}{n^s}~\forall n$
. For this $c(n)$ both the series in question converges absolutely for $Re(s)>1$.
So, $\left(\sum \limits_{n=1}^{\infty}\cfrac{1}{n^s}\right)\cdot \left(\sum \limits_{n=1}^{\infty}\cfrac{\mu(n)}{n^s}\right)=\sum \limits_{n=1}^{\infty}\cfrac{1*\mu(n)}{n^s}=1$ (as r9m pointed out in comments)
A: This is an expansion on Mahbub Alam's answer. I felt that a different presentation would make the explanation easier to understand.
Property $\boldsymbol{1}$
As long as each sum converges,
$$
\sum_{n=1}^\infty f\ast g(n)=\sum_{n=1}^\infty f(n)\sum_{n=1}^\infty g(n)
$$
because each factorization $n=ab$ accounts for $f(a)g(b)$ on each side.

Property $\boldsymbol{2}$
Multiplication by a Completely Multiplicative Function, $h$, distributes over Dirichlet Convolution:
$$
\begin{align}
(hf)\ast(hg)(n)
&=\sum_{d\mid n}h(d)f(d)h\left(\frac nd\right)g\left(\frac nd\right)\\
&=h(n)\sum_{d\mid n}f(d)g\left(\frac nd\right)\\
&=h(f\ast g)(n)
\end{align}
$$
That is, $(hf)\ast(hg)=h(f\ast g)$.

Property $\boldsymbol{3}$
Since $1$ and $\mu$ are both Multiplicative, so is $1\ast\mu$. For any power, $k\ge1$, of a prime, $p$:
$$
\begin{align}
1\ast\mu\left(p^k\right)
&=\overbrace{\ \ \ \ 1\ \ \ \ }^{p^0}\overbrace{\ \ -1\ \ \ \ }^{p^1}\overbrace{+0+0+\dots+0}^{p^j\text{ for }j\gt1}\\
&=0
\end{align}
$$
However, $1\ast\mu(1)=1$. Therefore,
$$
1\ast\mu(n)=\varepsilon(n)=\left\{\begin{array}{}
1&\text{if }n=1\\
0&\text{if }n\gt1
\end{array}\right.
$$

Conclusion
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n^s}\sum_{n=1}^\infty\frac{\mu(n)}{n^s}
&=\sum_{n=1}^\infty\frac1{n^s}\ast\frac{\mu(n)}{n^s}\tag{1}\\
&=\sum_{n=1}^\infty\frac{1\ast\mu(n)}{n^s}\tag{2}\\
&=\sum_{n=1}^\infty\frac{\varepsilon(n)}{n^s}\tag{3}\\[6pt]
&=1\tag{4}
\end{align}
$$
Explanation:
$(1)$: Property $1$
$(2)$: Property $2$
$(3)$: Property $3$
$(4)$: all but the $n=1$ term of $(3)$ are $0$
