Proving that $\sum{}_0^\infty z^{2^n} $ is unbounded when $z \rightarrow 1$ I understand boundedness of series this way: 


*

*A series is "unbounded" if the sequence of partial sums are unbounded.

*The sequence of partial sums is unbounded if I can prove that for an arbitrary bound $M_n$ on the sequence $A_n$  I can always find a bound larger than it ( $M_n < H_n$).


So if I want to show that  $\sum{}_0^\infty z^{2^n} $ is unbounded when z $\rightarrow 1 $, then I need show that the sequence of partial sums  $A_m =\sum{}_0^m z^{2^n}$ cannot be bounded when $z \rightarrow 1$. 
I just need some indication of what how to proceed as I am sure I am still missing something conceptually important.
 A: For all $m \in \mathbb{N}$
$$\sup_{z \in (1-\delta,1)}\sum_{n=0}^\infty z^{2^n} > \sup_{z \in (1-\delta,1)}\sum_{n=0}^m z^{2^n} >  \sup_{z \in (1-\delta,1)} (m+1)z^{2^m} = m+1$$
Thus, 
$$\limsup_{z \to 1-} \sum_{n=0}^\infty z^{2^n} = \lim_{\delta \to 0} \sup_{z \in (1-\delta,1)}\sum_{n=0}^\infty z^{2^n} > m+1$$
for all $m \in \mathbb{N}$, implying
$$\limsup_{z \to 1-} \sum_{n=0}^\infty z^{2^n} = +\infty, $$
and $S(z) = \sum_{n=0}^\infty z^{2^n} $ is unbounded. (There is a sequence $z_k \to 1$ such that $S(z_k) \to +\infty$.)
A: If
$mz < 1$
then
$(1-z)^m
\ge 1-mz$
(standard proof by induction).
Then,
if
$z > 1-1/2^{m+k}$
where $k \ge 2$
then
$\begin{array}\\
A_m 
&=\sum_{n=0}^m z^{2^n}\\
&\gt\sum_{n=0}^m (1-1/2^{m+k})^{2^n}\\
&\ge\sum_{n=0}^m (1-2^n/2^{m+k})\\
&=m+1-\sum_{n=0}^m 2^n/2^{m+k}\\
&=m+1-2^{-m-k}\sum_{n=0}^m 2^n\\
&=m+1-2^{-m-k}(2^{m+1}-1)\\
&>m+1-2^{1-k}\\
&> m\\
\end{array}
$.
Therefore
$A_m$
can can be made
arbitrarily large
by choosing $m$ large
and then choosing
 $z> 1-1/2^{m+2}$.
