Is there an expression for $\sum_{k=0}^{\infty} 2^{-(N^k)}$ in terms of an integer $N>1$? I was trying to solve for the limit as $n\rightarrow\infty$: $\displaystyle A_n=\sum_{i=1}^{n} a_i; a_i=\frac{2^{-i}}{i}$ and I landed at the inequality $\frac{N-1}{N}\left(S_{n+1}^{(N)}-a_{1}\right)<A_{N^{n+1}}-a_{1}<S_n^{(N)}$, for $N>2$; where $\displaystyle S_n^{(N)}=\sum_{k=0}^{n}N^k{a_{N^k}}=\sum_{k=0}^{n}2^{-(N^k)}$.
Whereas I have since found that $\displaystyle\lim_{n\to\infty} A_n=log2$ I am still curious about the series $S_n^{(N)}$
The series is convergent for all $N\in\{2,3,4,\ldots\}$
$$\sum_{k=0}^{\infty} 2^{-(N^k)}$$
Is there an expression for it in terms of $N$? If so, what is it and what are the steps to derive it?
 A: Let’s try to find a pattern within the series. One way is to use the Abel Plana formula. I will replace N with n for simplicity. The result has the Logarithmic Integral function similar to the gamma function. Note the real and imaginary parts are separated in the end. The complex argument function is also used here:
$$\mathrm{\sum_{k=0}^\infty 2^{-n^k}= \sum_{k=0}^\infty \left(\frac12\right)^{n^k}= \sum_{k=0}^\infty e^{{-n^k}\ln(2)}=\frac12 2^{-n^0}+\int_0^\infty 2^{-n^x} dx+i\int _0^\infty\frac{2^{-n^{ix}}-2^{-n^{-ix}}}{e^{2\pi x}-1}dx=\frac14-\frac{li\left(\frac12\right)}{ln(n)}+i\int _0^\infty \frac{1}{e^{2\pi x}-1}\left(2^{-e^{-x arg(n)} cos( x log(n))} cos(log(2) e^{-x arg(n)} sin(x log(n))) - 2^{-e^{x arg(n)} cos(x log(n))} cos(log(2) e^{x arg(n)} sin(x log(n)))+ i (sin(log(2) e^{-x arg(n)} sin( x log(n))) (-2^{-e^{-x arg(n)} cos(x log(n)))} - sin(log(2) e^{x arg(n)} sin(x log(n))) 2^{-e^{x arg(n)} cos( x log(n)))}\right)dx}$$
There may exist some other ways of evaluating the result, but this is an integral representation of it. Please correct me and give me feedback!
