# Asymptotic behavior of $\sum\limits_{n=0}^{\infty}x^{b^n}$ when $x\to1^-$

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 ?

• You may be interested in MSE question 2864987 "What's the limit of $\log_2(1-x) + x + x^2 + x^4 + x^8 ...$" which is a special case where $b=2$. – Somos Jan 29 at 14:38

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.