If $\sum\limits_{n=0}^\infty c_n$ is bounded uniformly, then $\lim\limits_{x\to1^-} \sum\limits_{n=0}^\infty c_nx^n$ exists? Determine whether the following holds:
If $\sum\limits_{n=0}^\infty c_n$ is bounded uniformly (i.e. there exists $M>0$ such that, for every $N$, $|\sum\limits_{n=0}^N c_n|\leq M$), then $\lim\limits_{x\to1^-} \sum\limits_{n=0}^\infty c_nx^n$ exists.
 A: The example
$$ a_n = \begin{cases} (-1)^k, & \text{if } n = 2^k \\ 0, & \text{otherwise} \end{cases} $$
indeed serves as a counter-example. As explained in the comment, this is a direct consequence of what is called the high-indices theorem. However, let me attempt a more elementary approach. Notice that
$$ f(x) = \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} (-1)^n x^{2^n} $$
converges (at least) for $|x| < 1$ and satisfies the identity
$$ f(x) = \sum_{n=0}^{N-1} (-1)^n x^{2^n} + (-1)^N f(x^{2^N}) $$
for any $|x| < 1$ and for any positive integer $N$. Now plug $x = 1 - 2^{-N}$ to this identity. Introducing the quantity $\alpha_N = (1 - 2^{-N})^{2^N}$, we find that
\begin{align*}
f(1 - 2^{-N})
&= (-1)^N f(\alpha_N) + \sum_{n=0}^{N-1} (-1)^n \alpha_N^{2^{n-N}} \\
&= (-1)^N \left[ f(\alpha_N) + \sum_{k=1}^{N} (-1)^k \alpha_N^{1/2^k} \right] \qquad \text{(where $k = N-n$)} \\
&= (-1)^N \left[ f(\alpha_N) + \sum_{k=1}^{N} (-1)^k + \sum_{k=1}^{N} (-1)^k \big(\alpha_N^{1/2^k} - 1 \big) \right].
\end{align*}
We notice that $\alpha_N \nearrow e^{-1}$ as $N\to\infty$ and satisfies $ \alpha_N^{1/2^k} = 1 + \mathcal{O}(2^{-k}) $, where the generic bound is independent of both $N$ and $k$. (Indeed, we can prove that $|\alpha_N^{1/2^k} - 1| \leq 2^{-k} \log 4$ for all $N$, $k$.) Thus by the Weierstrass $M$-test (or by the dominated convergence theorem, if allowed), the following convergence is guaranteed:
$$ \lim_{N\to\infty} \left[ f(\alpha_N) + \sum_{k=1}^{\infty} (-1)^k \big(\alpha_N^{1/2^k} - 1 \big) \right] = f(e^{-1}) + \sum_{k=1}^{\infty} (-1)^k \big(e^{-1/2^k} - 1 \big) =: S. $$
This is a rapidly converging series, and a numerical computation shows that
$$S \approx 0.500157922615\cdots.$$
A key observation is that this limit $S$ is not equal to $1/2$. (I am pretty sure that exponential speed of convergence is enough to provide a rigorous proof, but I will skip that.) Now since
$$ \lim_{m\to\infty} f(1-2^{-2m}) = S \qquad \text{and} \qquad \lim_{m\to\infty} f(1-2^{-(2m+1)}) = 1-S, $$
it follows that $f(x)$ does not converge as $x \to 1^-$.
