Let $q \in (0,1)$ and consider the following summation: $$S(q,n) = \sum_{i=1}^n {q^2}^i$$ Is there a closed form expression or upper and lower bounds for $S(q,n)$?
Specifically, I am looking for something like $$S(q,n) \approx \frac{q^2}{p_n(q)}$$ where $p_n(q)$ can be some polynomial of $q$.
I did some simulations and it seems it is possible to get such an expression. See here, for a plot of $q$ versus $S(q,n)$ for $n=10000$. The red curve was obtained using Matlab's rational fit function.
Notice that $$S(q,n) = \sum_{i=1}^n q^{2^i} = q^2\sum_{i=1}^n q^{2^i-2}$$ and thus $$p_n(q) \approx \left( \sum_{i=1}^n q^{2^i-2}\right)^{-1}.$$ The coefficients of powers of $q$ up to $q^{2^n-2}$ is the last expression coincide with those in the series: $$\left( \sum_{i=1}^\infty q^{2^i-2}\right)^{-1} = 1 -q^2 + q^4 -2q^6 + 3q^8 -4q^{10} + \dots$$ The coefficients (of even powers of $q$) in this series are given by http://oeis.org/A104977 As $p_n(q)$ one can take any truncation of this series as needed.
I am fairly confident there's no closed form expression for $S(q,n)$, at least not one that works for $S(q) = \lim_{n\to \infty}S(q,n)$.
The number $S(1/2 )$ is known as Kempner's number, and it was shown to be transcendental in 1916. So any closed-form algebraic expression for $S(q)$ would have to involve transcendental constants, and it's possible that the proof (in the above reference) could be adapted to show no closed-form expression exists.
