Asymptotic formula for an expression in terms of the second kind of stirling numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:24:54Z http://mathoverflow.net/feeds/question/117491 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117491/asymptotic-formula-for-an-expression-in-terms-of-the-second-kind-of-stirling-numb Asymptotic formula for an expression in terms of the second kind of stirling numbers liaomingxue 2012-12-29T09:16:37Z 2013-02-10T14:22:00Z <p>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.</p> <p>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 $1$. ?</p> http://mathoverflow.net/questions/117491/asymptotic-formula-for-an-expression-in-terms-of-the-second-kind-of-stirling-numb/117521#117521 Answer by Brendan McKay for Asymptotic formula for an expression in terms of the second kind of stirling numbers Brendan McKay 2012-12-29T14:51:09Z 2012-12-29T14:51:09Z <p>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. </p>