Mobius function and exp(2 pi i n x) We know that
\begin{equation}
\sum_{n \geq 1} \frac{\mu(n)}{n^{s}} = \frac{1}{\zeta(s)},
\end{equation}
and so, the left series can be plainly analytically continued to $\text{Re}(s) \leq 1$.
Recently, I have been interested in the following series
\begin{equation}
S(x, s) \equiv \sum_{n \geq 1} \frac{\mu(n)\exp[2 \pi i n x]}{n^{s}}.
\end{equation}
I have tried for a few weeks extending this $S(x, s)$ to the left half-plane $\text{Re}(s) \leq 1$. But I have now realized that this is a very hard problem to work with. 
In fact, I found out the work by T. Zhan and J.-Y. Liu which states that
\begin{equation}
A(x, \alpha) = \sum_{n \leq x} \mu(n) \exp[2 \pi i n \alpha] \ll x^{3/4 + \epsilon},
\end{equation}
assuming the Generalized Riemann hypothesis. That the evaluation of this partial
sum is connected with the existence of analytic continuation of S(x, s) must be clear to any guys being freak with the zeta-function.
I would like to know if, as a possible starter for tackling this extension problem, there has been any method known for analytically-continuing the series $\sum_{n \geq 1} \mu(n)/n^{s}$ into any neighborhood of the point s = 1 without using the property of the Riemann zeta-function. 
Or any results known about expressing the partial sums $A(x, \alpha)$ in terms of the zeros of some zeta-functions?
Thanks in advance.
Updated Part
Regarding the comments by Mr. Humphries:
I checked in your link the inequality 
\begin{equation}
\sum_{n \geq x} \mu(n)e^{2 \pi i n \alpha} \ll_{\alpha} x^{1/2 + \epsilon}.
\end{equation}
According to the Wikipedia, Carleson's theorem states that for
$L^{2}$-periodic function f, 
\begin{equation} 
\lim_{N \to \infty} \sum_{|n| \leq N} f^{*}(n)e^{i n x} = f(x)
\end{equation} 
for almost all $x \in \mathbb{R}$.
I have a difficulty verifying the $O(x^{1/2 + \epsilon})$-inequality with the suggested theorem.
It would be happier if I could see how this theorem is used. 
As you point out this inequality being "not quantitive", we can't know anything about the absolute constant depending on $\alpha$?
Update
@user1952009: Then, do you mean that the function 
\begin{equation}
F(\alpha, s) = \sum_{n \geq 1} \frac{\mu(n)e^{2 \pi i n \alpha}}{n^{s}}
\end{equation}
has a meromorphic continuation w.r.t $s$ for each fixed $\alpha \in [0, 1]$?
I am unsure of the statement that the limit of a sequence of meromophic functions is meromorphic. Is this statement a standard topic, or could I know of any reference? Could such a limit be even meaningful, if some members of the sequence have poles in an inappropriate way?
I actually have one more question, if you have the time. 
If $F(\alpha, s)$ can be extended meromorphically for each $\alpha$ as you described, then is there a way to show that for $\alpha$ near the points $0$ and $1$, $F(\alpha, s)$ takes finite values for s near the positive real axis? 
@Mr. Humphries: Is your function a member of $L^{2}[0, 2 \pi]$?
Update w.r.t. user1952009's answer
I have some elementary familiarity with sums of the form
\begin{equation}
\sum_{d|n}a(d)b(n/d).
\end{equation}
However, I could not figure out how 
\begin{equation}
\sum_{qm + a \geq x} \mu(qm + a) \ll x^{1/2 + \epsilon}
\end{equation}
implies
\begin{equation}
\sum_{n \geq x, \quad gcd(n, q) = 1} 
\mu(n)e^{2 \pi i n k /q} \ll x^{1/2 + \epsilon}.
\end{equation}
Does the latter sum come from the left sum of your equation, namely
\begin{equation}
\frac{1}{\phi(q)}\sum_{\chi (\mod q)} \frac{\overline{\chi}(a)}{L(s, \chi)} \quad ?
\end{equation}
Or is there any calculation which transforms $\sum \mu(qm + a)$ into $\sum \mu(n)\exp{2 \pi i k n / q}$? 
 A: the generalized Riemann hypothesis is for Dirichlet characters $\chi$. a Dirichlet character is multiplicative $\chi(nm) = \chi(n)\chi(m)$ and periodic $\chi(n+q) = \chi(n)$. these two facts show that $\chi(n) = 0$ whenever $gcd(n,q) > 1$, and $|\chi(n)| = 1$ otherwise, and finally that if $q$ is a prime power then $\chi(n) = e^{2 i \pi k \log_{g}^{(q)}(n)}$ where for each $k$ such that $gcd(k,q) = 1$ you get a different $\chi$ modulo $q$, and $\log_{g}^{(q)}(n)$ is the discrete logarithm modulo $q$ i.e.$g$ is a generator  of $\mathbb{Z}_q^\times$ and $g^{\log_{g}^{(q)}(n)} \equiv n \pmod{q}$.
we get the Dirichlet L-function $$L(s,\chi)  = \prod_p \frac{1}{1-\chi(p) p^{-s}} = \sum_{n=1}^\infty \chi(n) n^{-s}$$
which has a functional equation, like $\zeta$, and a Riemann hypothesis. 
now, the trick is that, for any $a$ such that $gcd(a,q) = 1$ :
$$\frac{1}{\phi(q)}\sum_{\chi \pmod{q}}  \overline{\chi(a)}\chi(n) = \left\{ \begin{array}{l}1 \text{ if } n \equiv a \pmod q \\ 0 \text{ otherwise }\end{array}\right.$$
which in fact reveals that a $\chi$ is its own discrete Fourier transform (and $\phi(q) = \sum_{n\le q, gcd(n,q) =1}1$  is the Euler totient).
finally :
$$\frac{1}{L(s,\chi)} = \sum_{n=1}^\infty \mu(n) \chi(n) n^{-s}$$
hence
$$\frac{1}{\phi(q)}\sum_{\chi \pmod q} \frac{\overline{\chi(a)}}{L(s,\chi)} = \sum_{m=0}^\infty \mu(qm+a) (qm+a)^{-s}$$
assuming the GRH for every $\chi$ modulo $q$, that series is holomorphic for $Re(s) > 1/2$, hence $$\sum_{qm+a \le x} \mu(qm+a) = \mathcal{O}(x^{1/2+\epsilon})$$
and we get if $k \in \mathbb{Z}$ and $gcd(k,q) = 1$ :
$$\sum_{n \le x, gcd(n,q) = 1} \mu(n) e^{2 i\pi n k/ q} = \mathcal{O}(x^{1/2+\epsilon})$$
and because of the multiplicative  properties of $\mu(n)$, this should prove that
assuming the GRH for every $\chi$ modulo $q$: 
$$\sum_{n \le x} \mu(n) e^{2 i\pi n k/ q}= \mathcal{O}(x^{1/2+\epsilon})$$
hence, asuming the GRH for every $\chi$ modulo every $q$ we get that it's still true when $k/q$ is irrational.
