This is not a proof but we give arguments for the following statement to hold
The sum
$$s = \sum_{i=0}^{\infty} c_{i} x^i$$
where $c_k \in [-1,+1]$ for $k=0,1,2,...$ converges only if $|x| < 1$.
Let us study the simple case of a periodic sign function, i.e. $c_{i+k}=c_i$ with $k=1,2,...$
Then for $k=2$ we have for the formally infinite sum (we write the more common $x$ instead of $\alpha$)
$$s_2 = \sum_{i=0}^{\infty} c_{i} x^i =\\c_0 + c_1 x + c_2 x^2 + c_3 x^3 + ... = \\c_0+ c_1 x + c_0 x^2 + c_1 x^3 ...=\\ (c_0 + c_1 x)(1+x^2 + x^4 + ...) \to \frac{(c_0 + c_1 x)}{1-x^2}$$
and for general $k$
$$s_k = \sum_{i=0}^{\infty} c_{i} x^i =(\sum_{i=0}^{k} c_i x^i)\left(1+x^k + x^{2k} + ...\right)=\frac{\sum_{i=1}^{k} c_i x^i}{1-x^k}$$
Hence the behaviour of the sum boils down to a finite sum times a geometric series. The convergence question is then easily answered and reads $ |x| \lt 1$.
For a non-periodic sign function this argument breaks down, of course. Hence it would be interesting to devise a non-periodic sign function and investigate that case.
Let us take the a non periodic function following the non-perodicity of a irrational number $z$.
Let $a_0, a_1, a_2, ...$ the binary representation of the number $z$.
Since we have $a_i \in [0,1]$, $c_i=2 a_i-1$ is a non-periodic sign function.
We now plot the sum $s$ as a function of $x$ for some famous irrational numbers



The emerging divergence at $|x| \to 1$ is manifest in all three cases.
This hints strongly towards a divergence of the sum for any distrubution of the signs when $|x| \to 1$.
In other words, the sum for abitrary $c_k$ converges only if $|x| \lt 1$.