This follow my previous post here, where Song has proven that $\forall b>1,\lim\limits_{x\to 1^{-}}\frac{1}{\ln(1-x)}\sum\limits_{n=0}^{\infty}x^{b^n}=-\frac{1}{\ln(b)}$, that is to say : $$\forall b>1,\sum\limits_{n=0}^{\infty}x^{b^n}=-\log_b(1-x)+o_{x\to1^-}\left(\log_b(1-x)\right)$$ (The $o_{x\to1^-}\left(\log_b(1-x)\right)$ representing a function that is asymptotically smaller than $\log_b(1-x)$ when $x\to1^{-}$, that is to say whose quotient by $\log_b(1-x)$ converges to $0$ as $x\to1^{-}$, see small o notation)
So we have here a first asymptotic approximation of $\sum\limits_{n=0}^{\infty}x^{b^n}$.
I now want to take it one step further and refine the asymptotic behaviour, by proving a stronger result which I conjecture to be true (backed by numerical simulations) : $$\sum\limits_{n=0}^{\infty}x^{b^n}=-\log_b(1-x)+O_{x\to1^-}\left(1\right)$$
(The $O_{x\to1^-}\left(1\right)$ representing a function that is asymptotically bounded when $x\to1^-$, see big o notation)
In other words, we want to go from :
"this sum is $-\log_b(1-x)$ + something that is asymptotically smaller than $\log_b(1-x)$ when $x\to1^-$"
to :
"this sum is $-\log_b(1-x)$ + something that is asymptotically bounded when $x\to1^-$"
And since $\log_b(1-x)$ diverges to $-\infty$ when $x\to1^-$, this is indeed a much more precise evaluation of the asymptotic behaviour !
Now, the way to go would be to show that $\sum\limits_{n=0}^{\infty}x^{b^n}+\log_b(1-x)$ is asymptotically bounded when $x\to1^-$, that is to say that $\exists M>0, \exists x_0\in\left(0,1\right) \text{ such that }\forall x\in\left[x_0,1\right), \left|\sum\limits_{n=0}^{\infty}x^{b^n}+\log_b(1-x)\right|\leqslant M$.
And to be honest, I'm kind of stuck. Any suggestion ?
Actually, this was almost done in my previous answer. I've shown that $$\begin{eqnarray} \sum_{n\ge 0}e^{-b^n \lambda}&=& -\frac{\ln \lambda}{\ln b}+O(1) \end{eqnarray}$$ where $e^{-\lambda}=x$. Thus $$ \sum_{n\ge 0}x^{b^n} =-\log_b (-\log x) +O(1). $$ It remains to show that $$ -\log_b (-\log x)=-\log_b (1-x)+O(1) $$ as $x\to 1^-$. This follows from $$ -\log_b (-\log x)+\log_b (1-x)=-\log_b\left(\frac{-\log x}{1-x}\right)\xrightarrow[]{x\to 1^-}0. $$ I hope this will help.