In an answer last year to another question, I implicitly asserted as an aside that that $\lim\limits_{r \to 1^-} \sum\limits_{k=0}^{\infty} (-1)^k r^{2^k}$ seemed empirically to be close to $\frac12$. I now think that there is not a limit, but that there are a limit inferior and a limit superior. So my question here is about finding $$\liminf\limits_{ r \to 1^{-}} \sum\limits_{k=0}^\infty (-1)^k r^{2^k} \text { and }\limsup\limits_{ r \to 1^{-}} \sum\limits_{k=0}^\infty (-1)^k r^{2^k}$$
Empirically, I now think
the limit inferior seems to be about $0.497250779$ and approached when $r$ is close to $1-\frac{0.6982}{2^{2n}}$ for increasing integers $n$
the limit superior seems to be about $0.502749221$ and approached when $r$ is close to $1-\frac{0.6982}{2^{2n+1}}$ for increasing integers $n$
If it helps, the terms of the series can be made positive with $\sum\limits_{k=0}^{\infty} (-1)^k r^{2^k} = \sum\limits_{m=0}^{\infty} r^{4^m}\left(1-r^{4^m}\right)$
Is there an analytical method for finding these values?
