Value of the sum $x^{2^n}$ for $0 < x < 1$ and $n$ a non-negative integer

Question

I'm interested in the following sum for $0 < x < 1$:

$ \sum_{n = 0}^{\infty} x^{2^n} $

Is there a known analytical solution? If not, how does one obtain the answer?

Answer 1

There is no analytical solution. However, you can find upper and lower bounds, which is essentially the next best thing. Call your sum $S$. You of course have the upper bound $S(x)<\frac{1}{1-x}$, $x\in[0,1)$. You also have the arbitrarily tight lower bound of $S(x)>\sum_{n=0}^N x^{2^n}$ for any non-negative integer $N$, which is most useful when $x\in[0,1) $ is small