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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: $S(.01,10^6) = 974.354$ , $S(.5,10^{10}) = 33.552$, $S(.3,10^2)=11.6682$

sheppa28
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  • What region of $(p,m)$ are you interested in? – Lee David Chung Lin Feb 04 '20 at 04:15
  • the series can be alternatively be written as $$S(p,m)=\sum_{k\geqslant 1}(1-\Pr[X\neq k]^m)$$ for a random variable $X$ with geometric distribution – Masacroso Feb 04 '20 at 06:43
  • @LeeDavidChungLin In most cases it'll be low p and high m. I tried taking the limit of p->0 and m->Inf with p*m->Lambda but the dependence on k dropped out yielding a an infinite sum. – sheppa28 Feb 04 '20 at 15:06
  • @Masacroso Correct. Now that I think about it, I should have actually worded it like that to begin with. It makes it more clear. – sheppa28 Feb 04 '20 at 15:08

2 Answers2

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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.

Claude Leibovici
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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 $G(a,b,m)=1-(1-a)^m+G(ab,b,m)$.

metamorphy
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