Upper bound for an alternating lacunary series I want to estimate an upper bound for the following alternating lacunary series $$\sum_{l=1}^\infty (-1)^{l+1}x^{2^l}$$ when $x$ is close to 1. In particular I want to show $$\limsup_{x\rightarrow 1-}\sum_{l=1}^\infty (-1)^{l+1}x^{2^l}<2/3.$$ Numerical analysis in Matlab shows the limit sup should be just a little higher than 0.5. 
In addition, in the paper "Some theorems concerning infinite series" by Hardy 1907 actually showed the limit does not exist because the function oscillates as $x$ goes to 1. However, I still can not find the upper bound that I want. Thanks!
 A: Let $f(x)=\sum_{l=1}^\infty (-1)^{l+1}x^{2^l}$, $0<x<1$. Since $f(x)+f(x^2)\equiv x^2$, it suffices to show that $\liminf_{x\to 1-} f(x)>1/3$. To this end, rewrite $f$ in the more convenient form 
$$
f(\sqrt{x}) = \sum_{l=0}^\infty (-1)^{l}x^{2^l}= \sum_{k=0}^\infty g\left(x^{4^k}\right) \tag{1}
$$ 
where $g(t)=t-t^2$. Since $g$ is positive on $(0,1)$, we can get a lower bound by  keeping only a few terms of the series (1); in fact, two will be enough.
For every $x\ge 7/8$ there exists a nonnegative integer $k$ such that $(7/8)^{4}\le x^{4^k}\le 7/8$. Therefore, it suffices to show that 
$$
g(t)+g(t^4)> \frac13,\quad (7/8)^{4}\le t\le 7/8 \tag{2}
$$
The verification of (2) is a tedious, but straightforward exercise with estimating the polynomial $h(t)=g(t)+g(t^4)=t-t^2+t^4-t^8$. 


*

*Exact calculation with rational numbers shows $h''((7/8)^4)<0$. 

*$h'''(t)=-24t(14t^4-1)<0$ for $t\ge (7/8)^4$, because $14\cdot (7/8)^{16}>1$.

*From 1 and 2 we see that $h''(t)<0$ when $t\ge (7/8)^4$.

*Hence, the minimum of $h(t)$ for $(7/8)^{4}\le t\le 7/8$ is attained at an endpoint. 

*Exact calculation with rational numbers shows that $h((7/8)^4)>\frac13$ and $h(7/8)>\frac13$.

Just as an illustration,  the plot of $h$ on the  interval $(7/8)^{4}\le t\le 7/8$.

