## Asymptotic formula for an expression in terms of the second kind of stirling numbers

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We have proved that the limit of $\sum_{k=0}^n r^2k^m / (1+r)^{k+1}$ when n approaches infinity is $\sum_{k=1}^m S(m,k)k!/r^{k-1}$ where S(m,k) is the second kind of stirling number.

Is there a simple asymptotic or approximate formula for the result $\sum_{k=1}^m S(m,k)k!/r^{k-1}$ with $m$ fixed and $r$ near $1$. ?

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## 1 Answer

For the question to make sense, you have to specify what the asymptotics is with respect to. For example, which variables are fixed and which are going to infinity. If $r>0$ is fixed and $m\to\infty$ (and probably in some other cases too), you are better off analyzing your initial sum rather than the Stirling version. The largest term is around $k= m/\ln(1+r)$ and the terms near that have a Gaussian shape with standard deviation $m^{1/2}/\ln(1+r)$. Euler-Maclaurin summation for the main part plus crude bounds for the tails will give it to you.

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A good suggestion. In fact and for our actual applications, $m$ is usuallly fixed and usually less than $10$, and $r$ will vary near 1, often not more than $10$ and not less than $1/2$. Therefore we do want an approximate result for $\sum_{k=1}^m S(m,k)k!/r^{k-1}$. – liaomingxue Dec 30 at 11:30

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So you have an exact expression with usually less than 10 terms. Why do you think there should be something better? – Brendan McKay Dec 30 at 14:03
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