Concrete Mathematics EXERCISE 9.46
Show that the Bell number $\varpi_n=e^{-1}\sum_{k\ge0}k^n/k!$ of exercise 7.15 is asymptotically equal to \[ m(n)^ne^{m(n)-n-1/2}/\sqrt{\ln n} \] where $m(n)\ln m(n) = n-\frac12$, and estimate the relative error in this approximation.
Part of the answer is that (According to the errata, I've edited the answer)
For convenience we write just $m$ instead of $m(n)$. By Stirling's approximation, the maximum value of $k^n/k!$ occurs when $k\approx m\approx n/\ln n$, so we replace $k$ by $m+k$ and find that \begin{align*} \ln\frac{(m+k)^n}{(m+k)!}=&n\ln m-m\ln m+m-\frac{\ln 2\pi m}2\\ &-\frac{(m+n)k^2}{2m^2}+O(k^3m^{-2}\log n)+O(1/m)\tag1 \end{align*} Actually we want to replace $k$ by $\lfloor m\rfloor+k$; this adds a further $O(km^{-1}\log n)$. The tail-exchange method with $|k|\le m^{1/2+\epsilon}$ now allows us to sum on $k$, ...
How can we derive equation (1) (especially when $|k|\le m^{1/2+\epsilon}$)? I try to expand $\ln(m+k)!$ using Stirling's approximation. It gets \[ \ln(m+k)!=(m+k)\ln(m+k)-(m+k)+\frac12\ln(m+k)+\sigma+O(1/m) \] where $e^\sigma=\sqrt{2\pi}$. However, the term $k\ln m$ in \[ (m+k)\ln(m+k)=(m+k)(\ln m+\ln(1+k/m))=(m+k)\left(\ln m-k/m+O(k/m)^2\right) \] never vanishes, and it's $\Omega(1)=\omega\left(k^3m^{-2}\log n\right)$ when $k$ is small.
Any help? Thanks!