One of our user asked: 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 asymptotical approximation of $\sum\limits_{n=0}^{\infty}x^{b^n}$ . I now want to take it one step further and refine the asymptotical 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 wh...
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