Closed-form or approximation to infinite sum $S(p,m) = \sum_{k=1}^\infty \left( 1 - {(1-(1-p)^{k-1}p)}^m\right)$ Is there a closed-form expression or good approximation to the following infinite sum?
$$S(p,m) = \sum_{k=1}^\infty \left( 1 - {(1-(1-p)^{k-1}p)}^m\right)$$
Was able to get numerical values when the upper bound is large enough, but can not figure out a closed-form formula to be used in general. Examples:  $S(.01,10^6) = 974.354$ , $S(.5,10^{10}) = 33.552$, $S(.3,10^2)=11.6682$
 A: This is not an answer but it is too long for a comment.
If ew assume that $p$ is small,we can write
$$S(p,m)=\sum_{n=0}^\infty m\,c_n^{(m)}\, p^n$$where $c_n^{(m)}$ is a polynomial of degree $n$ in $m$.
I have been able to establish the analytical expression of the $c_n^{(m)}$ up to $n=12$ but they are quite messy and I have not been able to see any pattern.
I give you below the very first ones
$$c_0=1$$
$$c_1=-\frac{m}{4}+\frac{1}{4}$$
$$c_2=\frac{m^2}{18}-\frac{7 m}{24}+\frac{17}{72}$$
$$c_3=-\frac{m^3}{96}+\frac{17 m^2}{144}-\frac{11 m}{32}+\frac{17}{72}$$
$$c_4=\frac{m^4}{600}-\frac{31 m^3}{960}+\frac{817 m^2}{4320}-\frac{229
   m}{576}+\frac{1291}{5400}$$
$$c_5=-\frac{m^5}{4320}+\frac{49 m^4}{7200}-\frac{1141 m^3}{17280}+\frac{2293
   m^2}{8640}-\frac{7681 m}{17280}+\frac{5153}{21600}$$
$$c_6=\frac{m^6}{35280}-\frac{71 m^5}{60480}+\frac{1711 m^4}{100800}-\frac{3797
   m^3}{34560}+\frac{24541 m^2}{72576}-\frac{1817 m}{3840}+\frac{101345}{442413}$$
 The next do not fit in a line.
If you want them, let me know.
A: We have $S(p,m)=G(p,1-p,m)$ where, for $|b|<1$, $$G(a,b,m)=\sum_{k=0}^\infty\big(1-(1-ab^k)^m\big)=\sum_{n=1}^m(-1)^{n-1}\binom{m}{n}\frac{a^n}{1-b^n};$$ the latter is obtained after expanding $(1-ab^k)^m$ (using the binomial formula) and summing over $k$.
In particular, we easily get the following limit (cf. comments to the OP): $$\lim_{m\to\infty}\frac{S(\lambda/m,m)}{m}=\sum_{n=1}^\infty\frac{(-\lambda)^{n-1}}{n\cdot n!},$$ with the RHS related to the exponential integral function.
For asymptotics of $G(a,b,m)$ with fixed $a$ and $b$, one can apply the Nørlund–Rice approach (like I did here). For $0<b<1$ and $b\leqslant a\leqslant 1$, the result is $$G(a,b,m)=\frac{1}{2}-\frac{H_m+\log a}{\log b}+\frac{1}{\pi}\sum_{n=1}^\infty\frac{1}{n}\Im\left[\exp\left(-2n\pi i\frac{\log a}{\log b}\right)\prod_{k=1}^m\left(1+\frac{2n\pi i}{k\log b}\right)^{-1}\right],$$ where $H_m=\sum\limits_{k=1}^m\frac1k$; this can be extended using the obvious $G(a,b,m)=1-(1-a)^m+G(ab,b,m)$.
