Showing $\displaystyle\lim_{z\rightarrow ^{-}1}\displaystyle\sum_{n\geq 0}z^{2^n}$ does not exist I have been trying to bound this below, as the TA suggested, by some taylor series of a function I know diverges at $x=1$, like $\log(\frac{1}{1-x})$ taylor expanded around zero: 
$$-\log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$$
Which obviously doesn't work for $x<1$. Any tips? I am also curious to see some different ways of proving this. 
 A: if $c_k \ge 0$ and $f(z) = \sum_{k=0}^\infty c_k z^k$ converges for $|z|< 1$ and $\sum_{k=0}^\infty c_k $ diverges, then $$\lim_{ r \to 1^-} f(r) = +\infty$$
simply write $$f_K(z) = \sum_{k=0}^K c_k z^k$$
then for every $K$ and $r \in [0,1[$ : $$f(r) \ge f_K(r)$$ hence for every $K$ :
$$\lim_{ r \to 1^-} f(r) \ge \lim_{ r \to 1^-} f_K(r) = \sum_{k=0}^K c_k$$
proving that $\lim_{ r \to 1^-} f(r)$ is unbounded.
A: From Cauchy's Condensation Test, we have
$$\begin{align}
\sum_{n=1}^\infty z^{2^n}&\ge \sum_{n=1}^\infty \frac{z^n}{n}\\\\
&=-\log(1-z) \tag 1
\end{align}$$
Therefore, we find from $(1)$ that 
$$\begin{align}
\lim_{z\to 1^-}\sum_{n=1}^\infty z^{2^n}&\ge -\lim_{z\to 1^-}\log(1-z)\\\\
&=\infty
\end{align}$$
And we are done!
A: What about the complex case? Define $$f(z)=\sum_{n=0}^\infty z^{2^n}\quad(|z|<1).$$
There have been several comments about how bad $f$ is at the boundary, but in fact it's worse than any of the commenters  seem to realize.
First, there does not exist a boundary point $e^{it}$ such that $\lim_{r\to 1^-}f(re^{it})$ exists. This follows from a "tauberian theorem" about "lacunary series", which shows that if $f$ has a radial limit at a given boundary point then the series itself converges at that point. Our series certainly converges  at no point of the boundary, hence it has a radial limit at no boundary point. (If I find a reference for that tauberian theorem I'll insert it here.)
Things are much worse than that. For example, although $f$ tends to infinity as you approach various boundary points radially, it does worse than that as you approach a boundary point in an arbitrary way from inside the disk.
In fact, for every $t$ there exists a sequence $(z_n)$ with $|z_n|<1$ and $z_n\to e^{it}$ such that the sequence $f(z_n)$ is dense in the plane!
We can say a little about why that's so.
For each $t\in \Bbb R$ let $S_t$ be the interior of the convex hull of $\{e^{it}\}\cup\{|z|<1/2\}$. The region $S_t$ is a "Stolz angle", or "nontangential approach region" - draw a picture to understand why approaching $e^{it}$ from within $S_t$ is known as "nontangential convergence".
Say $f$ has a nontangential limit at $e^{it}$ if $$\lim_{S_t\ni z\to e^{it}}f(z)$$exists. Say $f$ is nontangentially dense at $e^{it}$ if $f(S_t)$ is dense in the plane.
Plessner's Theorem If $g$ is holomorphic in the unit disc then for almost every $t$, either $g$ has a nontangential limit at $e^{it}$ or $g$ is nontangentially dense at $e^{it}$.
I haven't found an online reference for that. It's in various books, for example Garnett Bounded Analytic Functions, I think in Duren Theory of $H^p$ Spaces, various other books on Hardy spaces.
Now our function $f$ has a radial limit at no boundary point, hence a nontangential limit at no boundary point, so Plessaner says that $f$ is nontangentially dense at almost every boundary point. It follows easily from this that for every $t$ there exists $z_n\to e^{it}$ such that $f(z_n)$ is dense (note that $z_n$ tends to $e^{it}$, but not nontangentially).
A: If we choose $z$ real positive, and $z^{2^n}\ge 1/2$, then $z^{2^k}\ge 1/2$ for all $k\le n$. So taking account of the fact that the first term is $1$, our partial sum up to $n$ is $\gt n/2$, and so goes to $\infty$ as $n\to\infty$.
To make sure that $z^{2^n}\ge 1/2$, it is enough to choose $z\gt \exp(-\ln(2)/2^n)$. This is less than $1$.
A: If all you want is to show that the limit does not exist that's more or less obvious: $\sum z^{2^n}>\sum_{n=1}^Nz^{2^n}$, so the limit is larger than $N$ for every $N$.
But you can bound it below by that logarithm. Possibly interesting, also  it tells you how fast the function blows up. Note that $$\sum_{j=2^{n-1}+1}^{2^n}\frac{z^j}{j}\le z^{2^{n-1}}\sum_{j=2^{n-1}+1}^{2^n}\frac{1}{2^{n-1}}=z^{2^{n-1}}\quad(0<z<1).$$
