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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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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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  • $\begingroup$ 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 $\sum_{k=1}^m S(m,k)k!/r^{k-1}$. $\endgroup$
    – liaomingxue
    Commented Dec 30, 2012 at 11:30
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    $\begingroup$ So you have an exact expression with usually less than 10 terms. Why do you think there should be something better? $\endgroup$
    – Brendan McKay
    Commented Dec 30, 2012 at 14:03

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