Approximation of the sum involving binary entropy function Given the following sum:
$S(n) = \sum_{i=1}^{n} \frac{1}{(1-\operatorname{H}(p))^i}$
where $H$ is the binary entropy function defined as: 
$\operatorname{H}(p) = -p\log p - (1-p)\log (1-p) $.
Let $f(n) = \frac{n}{S(n)}$.
Assume $p$ is very small, is it possible to approximate the $S(n)$ and $f(n)$ defined above with simpler closed formulae e.g. a polynomial of $p$ without loosing much precision?
 A: The second order Taylor expansion of $H(p)$ may suffice for you. Expanding $H(p)$ around $p = 1/2$ it is not difficult to show that
$$H(p)=1-\frac{(1-2 p)^2}{2 \ln 2}+O\left(\left(p-\frac{1}{2}\right)^3\right)$$
Thus, using the approximation $1 - H(p) \approx \frac{(1-2 p)^2}{2 \ln 2}$, we get
$$S(n) \approx \sum_{i=1}^n \left(\frac{2 \ln 2}{(1-2 p)^2}\right)^{i},$$
which is has the simple closed form
$$S(n) \approx \frac{2 \ln 2 \left((1-2 p)^2\right)^{-n} \left(\left((1-2 p)^2\right)^n-(2 \ln 2)^n\right)}{4 p^2-4 p+1-2 \ln 2}.$$
The approximation should be good near $p = 1/2$, however in the numerical examples I have worked out, it seems pretty reasonable.
A: Almost eight years later
$$S(n) = \sum_{i=1}^{n} \frac{1}{\Big[1-\operatorname{H}(p)\Big]^i}=\frac{\Big[1-\operatorname{H}(p)\Big]^{-n}-1 }{\operatorname{H}(p) }$$
Answering this question, I proposed as a rather good approximation for the whole range of $p$
$$\operatorname{H}(p) \sim \big[4p(1-p)\big]^{\frac 34}\implies f(n)=\frac n {S(n)}=\frac{n\operatorname{H}(p) }{\Big[1-\operatorname{H}(p)\Big]^{-n}-1 }$$
