Proving that $\lim\limits_{x\to 1^{-}}\frac{1}{\ln(1-x)}\sum\limits_{n=0}^{\infty}x^{b^n}=-\frac{1}{\ln(b)}$ I conjecture that :
$$\forall b\in\mathbb{N}\setminus\lbrace0,1\rbrace,\lim\limits_{x\to 1^{-}}\frac{1}{\ln(1-x)}\sum\limits_{n=0}^{\infty}x^{b^n}=-\frac{1}{\ln(b)}$$
Which is well verified through numerical simulations.
Maybe I'm missing something obvious here, but I have absolutely no idea as of how to prove it. Uniform convergence is of course of no help here, the series $\sum\limits_{n=0}^{\infty}1$ being trivially divergent.
Any insight ?
 A: Let $x = e^{-\lambda}$, $\lambda \to 0^+$. We find that
$$
\sum_{n\ge 0}x^{b^n} =\sum_{n\ge 0}e^{-b^n \lambda}=\int_0^\infty e^{-b^t \lambda} \mathrm{d}t + \varepsilon_\lambda,
$$ where $|\varepsilon_\lambda |\le 1$ for all $\lambda>0$, i.e. $\varepsilon_\lambda =O(1)$. By making substitution $b^t\lambda =u$,
$$\begin{eqnarray}
\sum_{n\ge 0}e^{-b^n \lambda}&=&\frac{1}{\ln b}\int_\lambda^\infty e^{-u}\frac{\mathrm{d}u}{u}+O(1)\\&=&\frac{1}{\ln b}\int_\lambda^1 \frac{\mathrm{d}u}{u}+\frac{1}{\ln b}\int_\lambda^\infty \frac{e^{-u}-1_{\{u\le 1\}}}{u}\mathrm{d}u+O(1)\\
&=&-\frac{\ln \lambda}{\ln b}+O(1),
\end{eqnarray}$$ since $$\left|\int_\lambda^\infty \frac{e^{-u}-1_{\{u\le 1\}}}{u}\mathrm{d}u\right|\le\int_0^\infty \frac{|e^{-u}-1_{\{u\le 1\}}|}{u}\mathrm{d}u<\infty.$$ Finally, we have for all $b>1$,
$$\begin{eqnarray}
\lim_{x\to 1^-} \frac{1}{\ln(1-x)}\sum\limits_{n\ge 0}x^{b^n}&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{\ln \lambda+O(1)}{\ln(1-e^{-\lambda})}\\&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{\ln \lambda}{\ln(1-e^{-\lambda})}\\&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{1/\lambda}{e^{-\lambda}/(1-e^{-\lambda})}\\&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{e^\lambda(1-e^{-\lambda})}{\lambda}=-\frac{1}{\ln b}.
\end{eqnarray}$$
