TITLE: Ordinary Generating Function for Mobius
QUESTION [16 upvotes]: Is there any information known for the Ordinary Generating Function for Mobius?
$$
\sum_{n=1}^{\infty}  {\mu(n)}x^n
$$
I know that

It has radius of convergence 1.
Does not have limit as $x\rightarrow 1$. 

My question is 

Does it have limit as $x\rightarrow \textrm{exp}(i\theta)\neq 1$?
Is there a similar result for $\textrm{exp}(i\theta)\neq 1$ with 
$$
\sum_{n\leq x}\mu(n)=o(x)$$  i.e. is there a result of this type?
$$
\sum_{n\leq x}\mu(n)\textrm{exp}(in\theta) = o(x)$$

EDIT1: Aside from my questions, there are some known results for the OGF for Mobius. 
EDIT2: Found a reference for (III)
Reference: Jason P. Bell, Nils Bruin, Michael Coons "Transcendence of generating functions whose coefficients are multiplicative" available at http://www.ams.org/journals/tran/2012-364-02/S0002-9947-2011-05479-6/
P. Borwein, T. Erd´elyi, and F. Littman, Polynomials with coefficients from a finite set
available at http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P195.pdf
(I) This function has the unit circle as natural boundary. 
(II) This function is a transcendental function. 
(III) It is not bounded on any open sector($\{z:|z|<1, z=re^{i\theta}, \alpha<\theta<\beta\}$)of unit disc. 
This result (III) together with Baire Category answers my question 1 with a dense set of $\theta$'s, but as Prof Tao mentioned, there can be a possibility that it is bounded for some values of $\theta$.

REPLY [14 votes]: Mobius randomness heuristics suggest that $\sum_n \mu(n) (r e^{i\theta})^n$ does not converge to a limit as $r \to 1^-$ for any (or at least almost any) $\theta$.  If it did converge for some $\theta$, then we would have
$$\sum_n \mu(n) e^{in\theta} \psi_k(n) \to 0$$
as $k \to \infty$, where
$$ \psi_k(n) := (1-2^{-k-1})^n - (1-2^{-k})^n.$$
The cutoff function $\psi_k$ is basically a smoothed out version of the indicator function $1_{[2^k,2^{k+1}]}$.  The Mobius randomness heuristic (discussed for instance in this article of Sarnak) then suggests that the sum $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ should have a typical size of $2^{k/2}$, and so should not decay to zero as $k \to \infty$.
Note from Plancherel's theorem that the $L^2_\theta$ mean of $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ is indeed comparable to $2^{k/2}$.  This does not directly preclude the (very unlikely) scenario that this exponential sum is very small for many $\theta$ and only large for a small portion of the $\theta$, but if one optimistically applies various versions of the Mobius randomness heuristic (with square root gains in exponential sums) to guess higher moments of $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ in $\theta$, one is led to conjecture a central limit theorem type behaviour for the distribution of this quantity (as $\theta$ ranges uniformly from $0$ to $2\pi$), i.e. it should behave like a complex gaussian with mean zero and variance $\sim 2^k$, and in particular it should only be $O(1)$ about $O(2^{-k})$ of the time, and Borel-Cantelli then suggests divergence for almost every $\theta$ at least.  Unfortunately this is all very heuristic, and it seems difficult with current technology to unconditionally rule out a strange conspiracy that makes $\sum_n \mu_n (re^{i\theta})^n$ bounded as $r \to 1$ for some specific value of $\theta$ (say $\theta = \sqrt{2} \pi$), though this looks incredibly unlikely to me.  (One can use bilinear sums methods, e.g. Vaughan identity, to get some nontrivial pointwise upper bounds on these exponential sums when $\theta$ is highly irrational, but I see no way to get pointwise lower bounds, since $L$-function methods will not be available in this setting, and there may well be some occasional values of $\theta$ and $k$ for which these sums are actually small.  But it is likely at least that the $\theta=0$ theory can be extended to rational values of $\theta$.)