What is the sum of the series $\sum \limits_{k=0}^\infty q^{2^k}$ if $|q|\lt1$ I wonder if anybody knows how to calculate the series below?
$\sum \limits_{k=0}^\infty q^{2^k}$ if $|q|\lt1$?
Thanks a lot for answers.
 A: Thanks a lot for links
$\sum_{k=0}^\infty q^{2^k}$ if $|q|\lt1$
I also have noticed that the series above is related to the  function equation
$G(2x)-G(x)=e^x-1$.
$$G(x)=\sum_{k=1}^\infty \frac{x^k}{k!(2^k-1)}$$
$\lim_{n\to\infty} (n+1+G(x.2^{n+1})-G(x)) = 
\ e^x+e^{2x}+e^{4x}+e^{8x}+\dots$
where $q=e^x$
And also other relation of that series,
$e^x+e^{2x}+e^{4x}+e^{8x}+\dots=F(x)$ where $q=e^x$
$x=2^t$
$e^{2^t}+e^{2^{t+1}}+e^{2^{t+2}}+\dots=F(2^t)=H(t)$
$e^{2^t}+H(t+1)=H(t)$
$H(t+1)-H(t)=-e^{2^t}$
$H(t)= k-\int_0^t e^{2^t} \mathrm{d}t+ \frac12 e^{2^t}-\frac1{12} \frac{\partial}{\partial t}(e^{2^t})+\dots$
Here need to find $k$ to complete the function $H(t)$ ( constants are bernoulli numbers)
Please advice me if anything wrong in my relations and need a hint to find $k$.
In other way, I found the way to find  exact solution via Fourier transfom
$H(t+1)-H(t)=-e^{2^t}$  //We need to take fourier integral both side
$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t-\int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t=-\int_{-\infty}^{+\infty} e^{2^{t}}e^{-2πift} \mathrm{d}t$
$V(f)= \int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t$
$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t=\int_{-\infty}^{+\infty} H(z)e^{-2πif(z-1)} \mathrm{d}z=V(f).e^{2πif}$
$e^{2πif}V(f)-V(f)=-\int_{-\infty}^{+\infty} e^{2^{t}}e^{-2πift} \mathrm{d}t$
$V(f)=-\int_{-\infty}^{+\infty} (e^{2^{t}}e^{-2πift})/(e^{2πif}-1) \mathrm{d}t$ 
//Now We need to take inverse fourier transform
$H(z)=\int_{-\infty}^{+\infty} V(f) e^{2πifz} \mathrm{d}f=\int_{-\infty}^{+\infty}  e^{2πifz}(-\int_{-\infty}^{+\infty} (e^{2^{t}}e^{-2πift})/(e^{2πif}-1) \mathrm{d}t).\mathrm{d}f $
$H(z)=\int_{-\infty}^{+\infty}  \int_{-\infty}^{+\infty} e^{2πifz} (e^{2^{t}}e^{-2πift})/(1-e^{2πif}) \mathrm{d}t\mathrm{d}f $
Exact solution of H(t) in integral representive way
$\sum_{k=0}^\infty q^{2^k}=H(t)=H(log_2(lnq))$ 
