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Hi, is it possible to give an explicit formula for the function G(s) defined for positive s as

$G(s) := \lim_{N\to\infty} \sum_{k=1}^N \frac{1}{k}{N\choose k} \left(\frac{s}{N}\right)^k \left(1-\frac{s}{N}\right)^{N-k}$.

Wolfram Mathematica says the sum for finite $N$ is some hypergeometric function, the limit of it as $N\to\infty$ is not able to express explicitely.

Thanks, Ian

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2 Answers

up vote 6 down vote accepted

Unless I've made some horrible miscalculation your limit is the same as $$\lim_{N\to \infty}\left(1-\frac{s}{N}\right)^N \int_{0}^{\frac{s}{N-s}}\frac{(1+x)^N-1}{x} \ dx$$ which is equal to $$e^{-s} \lim_{N\to \infty} \int_{0}^{\frac{Ns}{N-s}}\frac{\left(1+\frac{y}{N}\right)^N-1}{y} \ dy=e^{-s}\int_{0}^s \frac{e^x-1}{x} \ dx.$$ And so you get the final answer $$e^{-s}\left(Ei(s)-\gamma -\log s\right)$$ where $Ei(x)$ is the Exponential integral and can not be "simplified" further.

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Here's a probabilistic solution to your problem. Suppose $X_N$ is a binomial random variable with parameters $N$ and $s/N$ (ie. $X_N$ counts the number of heads in $N$ independent coin flips, each with probability $s/N$ of heads). Also let $P$ be a Poisson random variable with mean $s$, so that for all non-negative integers $k$: $$Prob[P=k] = e^{-s}\frac{s^k}{k!}.$$ A well-known result sometimes called the law of rare events implies that the distribution of $X_N$ converges to that of $P$ as $N\to +\infty$. In particular, for any bounded real-valued function $f$ defined on the non-negative integers: $$E(f(X_N)) = \sum_{k=0}^N \binom{N}{k}\left(1-\frac{s}{N}\right)^{N-k}\left(\frac{s}{N}\right)^{k}f(k)\to E(f(P))=\sum_{k=0}^{+\infty}e^{-s}\frac{s^k}{k!}f(k).$$

Apply this to $f$ satisfying $f(x)=1/x$ if $x>0$, $f(0)=0$, and the LHS becomes your expression. The RHS becomes: $$\sum_{k=1}^{+\infty}e^{-s}\frac{s^k}{k!\times k} = e^{-s}\int_{0}^{s}\frac{e^{u}-1}{u}du,$$ which I guess is the same as the previous answer.

Added note: a quantitative version of the law of rare events gives the error bound: $$\forall f:N\to [0,1], |E(f(X_N))-E(f(P))|\leq s\left(1-e^{-s/N}\right);$$ this allows for simultaneous limits in N and $s$, and goes to $0$ iff $s^2/N\to 0$.

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