TITLE: Sum of subset of geometric series: a^2^n
QUESTION [8 upvotes]: The formula for 1 + a + a^2 + .... where 0 < a < 1 is $\frac{1}{1-a}$, but what if you wanted to sum only those where the exponent is a power of 2? That is,
$S = a + a^2 + a^4 + a^8 + \cdots$ 
I feel like this is an easy one but I just can't seem to find a closed expression for it, nor search for it on Google.

REPLY [16 votes]: Mahler proved in the 1930s that the values of $f(z)=\sum_{n=0}^\infty z^{d^n}$, $d>1$ is an integer, are transcendental for any algebraic $z$ satisfying $0<|z|<1$. A related problem of transcendence of the function $f(z)$ was discussed in this question. This motivates nonexistence of simple formula like $1/(1-z)$ for $f(z)$.