<s>[INST] Upper bound for a series $\sum_{n=0}^{\infty}2^n r^{2^n}$ for $r\in (0,1)$ It is clear that for $r\in (0,1)$ fixed, the series $$f(r)=\sum_{n=0}^{\infty}2^n r^{2^n}$$ is convergent, and also it is not to hard show that $f(r)$ is not uniformly convergent in $(0,1)$. Since $\sum_{n=0}^{\infty}r^n=\frac{1}{1-r}$, I think this sum function $f(r)$ may be similar.
So can I find some upper bound such as $\frac{1}{1-r^\alpha}$ to dominate $f(r)$? Sincerely thanks!
 [/INST] This is a different approach with an arbitrary good bound for $x\in(0,1)$.
$$f(x)=\sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty 2^{n-1} x^{2^n} = \sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty \underbrace{x^{2^n} + ... + x^{2^n}}_{2^{n-1} \text{ terms}} \\
\leq \sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty x^{2^{n-1}+1}+x^{2^{n-1}+2}+...+\underbrace{x^{2^{n-1}+2^{n-1}}}_{x^{2^n}} \\
=\sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=2^N+1}^\infty x^n = \sum_{n=0}^N 2^n x^{2^n} + \frac{2\,x^{2^N+1}}{1-x} \, .$$
For example, for $N=1$ you have $$f(x)\leq \frac{x(x+1)}{1-x} \, .$$
Aside from that, you may want to check these posts

*

*What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$.

*Evaluate $\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right)-\log_2\frac{1}{1-x}\right)$
which make statements about the behaviour of your function in the vicinity of $x=1$.
Since $$f(x)=\sum_{n=0}^\infty 2^n x^{2^n} = x \frac{{\rm d}}{{\rm d}x} \underbrace{\sum_{n=0}^\infty x^{2^n}}_{g(x)}$$
you may want to study $g(x)$ instead. $g(x)$ (and $f(x)$ alike) is obviously strictly increasing on $(0,1)$. Thus you may seek the behaviour as $x\rightarrow 1$. The first link shows that the function $$G(x)=g(x)+\log_2(1-x)$$ has a mean limit of $$\overline{G(1)}=\overline{\lim_{x\rightarrow 1^-} G(x)} = \frac{1}{2} - \frac{\gamma}{\log 2} \, .$$
However, $G(x)$ does not converge as $x\rightarrow 1$. If you write down the series expression for $G(1)$, you'll find it is still depending on $x$ and in the variable $t$, where $x=e^{-2^{-t}}$, it has a period of $1$. In the second link I calculate the fourier-expansion of $G$ in the variable $t$ with the result
$$G(1)=\frac{1}{2}-\frac{\gamma}{\log 2} + \sum_{n=1}^\infty A_n \cos\left(2\pi nt- \varphi_n\right)$$
where
\begin{align}
A_n&=\frac{2\left|\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right|}{\log 2}=\sqrt{\frac{2}{n\sinh\left(\frac{2\pi^2n}{\log 2}\right)\log 2}} \\
-\varphi_n&=\arg \left( \Gamma\left(\frac{2\pi i \,n}{\log 2}\right) \right) \, .
\end{align}
Notice that the first harmonic is already a very good approximation, $A_2/A_1<10^{-6}$.
Thus in the vicinity of $x=1^-$ you can write
$$g(x)=G(1)-\log_2(1-x) \\
f(x)=x \frac{{\rm d}}{{\rm d}x} g(x) = \frac{1}{\log 2} \frac{x}{1-x} - \frac{2\pi}{\log 2} \frac{-1}{\log x} \sum_{n=1}^\infty n A_n \sin\left(2\pi nt - \varphi_n \right) \\
\leq \frac{1}{\log 2} \frac{1}{1-x} \left( x + 2\pi \sum_{n=1}^\infty n A_n \right) < \frac{x+10^{-5}}{\log 2} \frac{1}{1-x}$$
with $\frac{-1}{\log x} \leq \frac{1}{1-x}$ for $0<x<1$.
</s>