Does the family of series have a limit? For $r<1$ define $F(r)=\sum_{n\in\mathbb N}(-1)^nr^{2^n}$. Does $F$ have a limit as $r\nearrow 1$?
 A: Note that
$$
F(r)=r-F(r^2)\tag{1}
$$
Thus, if $a=\lim\limits_{r\to1^-}F(r)$ exists, then
$$
a=\lim_{r\to1^-}F(r)=\lim_{r\to1^-}r-\lim_{r\to1^-}F(r^2)=1-a\tag{2}
$$
Therefore, if the limit exists then it is $a=\frac{1}{2}$.
Applying equation $(1)$ twice, we get
$$
F(r)=r-r^2+F(r^4)\tag{3}
$$
As $r\to1$, $(3)$ indicates $F$ tends toward being periodic in $-\log(-\log(r))$ with period $\log(4)$. Note that as $r\to1^-$, $-\log(-\log(r))\to\infty$.  $F(r)$ is the sum of the lengths of the intervals in the following animation

The value of the sum oscillates between $0.49728$ and $0.50272$ over each period. Therefore, $\lim\limits_{r\to1^-}F(r)$ does not exist.
A: My question was connected with this one. Namely, consider the sequence $(1, -1, -1, 1, 1, 1, 1, -1, \dots)$, where $(-1)^{k}$ stands for indices from $2^{k-1}$ to $2^k-1$. The Cesaro means can easily be calculated and they don't have a limit. The function $F$ here corresponds to the Abel means, and the equivalence of these summation methods for the bounded sequence implies that the Abel means diverge, too.
