[Community Question] Calculus: Asymptotic behavior of $\sum\limits

[Community Question] Calculus: Asymptotic behavior of $\sum\limits_{n=0}^{\infty}x^{b^n}$

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 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 asymptotical 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 ?

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).

Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$

Let $\mathbb{Z}(p^{\infty})$ be defined by $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$ , I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$ . i try several ways but I have not been successful, some help ?? thank you

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