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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 ?

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  • $\begingroup$ A stronger result is also true. $\sum_{n=0}^{\infty} x^{b^n}+\frac{\ln(1-x)}{\ln b}$ is boundedly oscillating as $x\rightarrow 1-$. $\endgroup$ Jan 29, 2019 at 7:49

1 Answer 1

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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 $b>1$, $$\begin{eqnarray} \lim_{x\to 1^-} \frac{1}{\ln(1-x)}\sum\limits_{n\ge 0}x^{b^n}&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{\ln \lambda+O(1)}{\ln(1-e^{-\lambda})}\\&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{\ln \lambda}{\ln(1-e^{-\lambda})}\\&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{1/\lambda}{e^{-\lambda}/(1-e^{-\lambda})}\\&=&-\frac{1}{\ln b}\lim_{\lambda \to 0^+}\frac{e^\lambda(1-e^{-\lambda})}{\lambda}=-\frac{1}{\ln b}. \end{eqnarray}$$

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  • $\begingroup$ Thank you, great answer ! $\endgroup$ Jan 29, 2019 at 12:00
  • $\begingroup$ Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more. $\endgroup$ Jan 29, 2019 at 13:30
  • $\begingroup$ @HarmonicSun Thank you for information :) I've left my answer to your new question ! $\endgroup$ Jan 29, 2019 at 13:43

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