Does $z+z^2+z^4+z^8 + .... $ ever become unboundedly negative? I was analyzing the function 
$$ \Delta(x)  = \sum_{k=0}^{\infty} x^{2^k} $$ 
Over $x \in \mathbb{C}  \text{ s.t. } |x| \le 1$.
What I'm trying to answer is: does there exist a sequence of complex numbers $x_i$ such that $|x_i| \le 1$ and $ \forall \delta < 0  \\ \exists \  \tau 
 \ | \  \forall \ n > \tau \    \Delta(x_n) < \delta  $
I.E. a sequence in the unit disk whose limit value under this function is $-\infty$. 
I'm finding that travelling from 0 towards any fixed point in unit circle (where there are known singularities) isn't fruitful, since it always tends towards positive infinity. 
My (rather superficial) argument is as follows, consider any such direction $a+bi$ where $a^2 + b^2 = 1$. Observe that for any choice of distance $\epsilon$ exists an arbitrarily close $u+vi$ such that $u+vi$ is the solution to $x^{2^j} = 1$ for some set of positive $j$ and therefore $\Delta(u+vi) = +\infty$. 
So if $\Delta(a+bi)=-\infty$ then for any positive $\epsilon$ one can find a point on the unit disk where $\Delta(u+vi) = +\infty$. 
So rigorously this means nothing. But intuitively, the function needs to be break continuity on its disk (which it already has) but in some weaker sense, I wonder if that has implications on the convergent series (to the respective points).
All in all, I'm a bit stuck. 
 A: I suspect the nearest you are going to get to this is something like $$z=-\frac12 + i \frac{\sqrt{3}}{2}$$ which has $z^2=\bar z, z^4=\bar{\bar{z}}=z,z^8=\bar{\bar{\bar{z}}}=\bar z, \ldots$, i.e. $z^{2^k} = -\frac12 + i \frac{(-1)^k\sqrt{3}}{2}$, where the partial sum $S_n=\sum\limits_{k=0}^{n} z^{2^k}= -\frac{n+1}2 + i \frac{\sqrt{3}}{4}\left(1+(-1)^n\right)$ is close to $-\frac {n+1}2$ in the sense that $\left|\frac{S_n- \left(-\frac {n+1}2\right)}{-\frac {n+1}2}\right| \to 0$ as $n$ increases, and $-\frac n2 \to -\infty$ as $n$ increases.  Another way of looking at this is that the magnitude of $S_n$ increases without limit and the argument of $S_n$ converges to $\pi$. 
This sidesteps the issue that $S_n$ is not real when $n$ is even though it is when $n$ is odd, and that imaginary part of $S_n$ does not converge.  
In your formulation, consider $x = re^{i 2\pi/3}$ with $r$ real and $0 \lt r \lt 1$: 
$$\sum\limits_{k=0}^{\infty} x^{2^k} = -\frac12\sum\limits_{k=0}^{\infty} r^{2^k} + i\frac{\sqrt{3}}{2}\sum\limits_{k=0}^{\infty} (-1)^k r^{2^k} $$
and the real part of this is not an issue as $\lim\limits_{r \to 1^-} \left[-\frac12\sum\limits_{k=0}^{\infty} r^{2^k}\right] = -\infty$, but I think the imaginary part may be  if $\lim\limits_{r \to 1^-} \left[i\frac{\sqrt{3}}{2}\sum\limits_{k=0}^{\infty} (-1)^k r^{2^k}\right]$ is not zero (empirically it seems to be close to $i\frac{\sqrt{3}}{4}$ but I am not confident about this).  
