Proving $\sum\limits_{n=1}^{\infty } \frac{\mu (n)}n=0$ where $\mu$ is the Möbius function 
How to prove that Dirichlet series $$\sum _{n=1}^{\infty } \frac{\mu (n)}{n^s}$$ converges to $0$ at $s=1$, where $\mu(n)$ is the Möbius function?

I know it is equal to reciprocal of Riemann zeta function $\frac{1}{\zeta (s)}$ and can be also represented by Euler product $\prod _{n=1}^{\infty } \left(1-p_n^{-s}\right)$, but the product is called divergent if its value goes to 0 and none of its factors is equal to zero, so I think this product will not help me.
 A: It is easy to see that
\[\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = 0,\]
because the left-hand side is equal to $1/\zeta(s)$ for $\Re(s) > 1$, and $\zeta(s)$ has a pole at $s = 1$. However, the fact that this limit is equal to $0$ does not imply that
\[\sum_{n \leq x} \frac{\mu(n)}{n} = o(1).\]
This is much harder to prove, as has been noted in the comments; you need to additionally know that $\zeta(s)$ is nonvanishing on the line $\Re(s) = 1$.
There are several approaches to prove this result. One can prove this via Perron's inversion formula to show that this is equal to
\[\frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \frac{1}{\zeta(s + 1)} \frac{x^s}{s} \, ds\]
for $\sigma > 0$, then move the contour to the line $\sigma = - c/\log x$ for some $c > 0$ and use the highly nontrivial fact that $\zeta(s)$ has a zero-free region, in which one can bound it from below.
Another approach is to show that $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$ is equivalent to the bound $\psi(x) \sim x$, where $\psi(x) = \sum_{p^k \leq x} \log p$. This can be done via "elementary methods" in about half a page of work. Of course, the statement that $\psi(x) \sim x$ is equivalent to the prime number theorem via partial summation.
