Finding the sum- $x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

Solution 1:

I have worked on this series before.

There is no simple closed form as a geometric series has.

(such as $x+x^2+x^3+x^4+...=\frac{x}{1-x}$ where $|x|<1$).

You can see more information in the link about Lacunary function.

The series can be expressed in closed form of double integral. I shared my result below.

$x+x^{2}+x^{4}+x^{8}+\dots=F(x)$

Let's transform $x=e^{-2^t} \tag{1}$

Where $-\infty<t<\infty$ Thus x will be in $ (0,1)$ and $F(x)$ is not divergent in this range.

$e^{-2^t}+e^{-2^{t+1}}+e^{-2^{t+2}}+\dots=F(e^{-2^t})=H(t)$

$e^{-2^t}+H(t+1)=H(t)$

$H(t+1)-H(t)=-e^{-2^t}$

The Fourier transform of both sides

$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t-\int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$

$$V(f)= \int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t$$

$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t=\int_{-\infty}^{+\infty} H(z)e^{-2πif(z-1)} \mathrm{d}z=V(f)e^{2πif}$$

$$e^{2πif}V(f)-V(f)=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$

$$V(f)=\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t$$

Now we need to take the inverse Fourier transform

$$H(z)=\int_{-\infty}^{+\infty} V(f) e^{2πifz} \mathrm{d}f=\int_{-\infty}^{+\infty} e^{2πifz}\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f $$

The closed form of $H(z)$ in integral expression: $$H(z)=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{2πifz} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f $$

$$\sum_{k=0}^\infty x^{2^k}=H(\log_2(-\ln x))=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{2πif\log_2(-\ln x)} \frac{e^{-2^{t}-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f$$

Where $0<x<1$

$$\sum_{k=0}^\infty x^{2^k}= \int_{-\infty}^{+\infty} \frac{e^{2πif\log_2(-\ln x)}}{1-e^{2πif}} \int_{-\infty}^{+\infty} e^{-2^{t}-2πift} \mathrm{d}t\,\mathrm{d}f=\int_{-\infty}^{+\infty} e^{-2^{t}} \int_{-\infty}^{+\infty} \frac{e^{2πif(\log_2(-\ln x)-t)}}{1-e^{2πif}} \mathrm{d}f\,\mathrm{d}t$$

Note: I made the update on 07/29/2016 . Variable change was $x=e^{2^t}$ at Tag (1) and it had problems as @leonbloy 's comment below. Thanks for the comment. Please let me know if you notice something else in definitions.