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 $$S(q,n) \approx \frac{q^2}{p(q)}$$ where $p(q)$ 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.

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.

Let $q \in (0,1)$ and consider the following summation: $$S(q) = \sum_{i=1}^n {q^2}^i$$ Is there a closed form expression or upper and lower bounds for $S(q)$ (preferably for finite $n$ but I am ok with the infinite summation as well) ?

Specifically, I am looking for something like $$S(q) \approx \frac{q^2}{p(q)}$$ where $p(q)$ 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 $S(q)$ for $n=10000$. The red curve was obtained using Matlab's rational fit function.

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 $$S(q,n) \approx \frac{q^2}{p(q)}$$ where $p(q)$ 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.