Asymptotics for Bell number 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!
 A: You're doing the right things, as far as I can tell.  Maybe you're not using $n = m \log m + 1/2$ or $n/\log n = m + o(m)$ in the right place?  (The latter asymptotic comes from $\log n = \log m + \log(\log m + 1/(2m))$ and doing the division of $n$ by $\log n$.  The dominant term is $m$.)   At any rate, here's the derivation I get.

We need Stirling's approximation
$$\log (m+k)! = (m+k) \log(m+k) - (m+k) + \frac{1}{2}\log(m+k) + \frac{1}{2} \log (2\pi) + O\left(\frac{1}{m}\right)$$ 
and $$\log(m+k) = \log m + \log\left(1 + \frac{k}{m}\right) = \log m + \frac{k}{m} - \frac{k^2}{2m^2} + O\left(\frac{k^3}{m^3}\right).$$
We have 
$$\log \frac{(m+k)^n}{(m+k)!} = n \log (m+k) - \log(m+k)!.$$
Let's take these two terms separately.
\begin{align*}
n \log (m+k) &= n \log m  + \frac{nk}{m} - \frac{nk^2}{2m^2} + O\left(\frac{nk^3}{m^3}\right) \\
&= n \log m + k \log m + \frac{k}{2m} - \frac{nk^2}{2m^2} + O\left(\frac{k^3 \log n}{m^2}\right). \tag{1}
\end{align*}
And 
\begin{align}
- \log(m+k)! = &-(m+k)\left(\log m + \frac{k}{m} - \frac{k^2}{2m^2} + O\left(\frac{k^3}{m^3}\right)\right) \\
& + (m+k) - \frac{1}{2}\left(\log m + \frac{k}{m}  + O\left(\frac{k^2}{m^2}\right)\right) - \frac{1}{2} \log (2\pi) + O\left(\frac{1}{m}\right) \\
=&- (m+k)\log m -k - \frac{k^2}{m} + \frac{k^2}{2m} + m+k - \frac{\log m}{2} - \frac{k}{2m}  \\
&- \frac{1}{2} \log (2\pi) + O\left(\frac{k^3}{m^2}\right) +  O\left(\frac{1}{m}\right)\\
=& - (m+k) \log m + m - \frac{\log(2 \pi m)}{2} - \frac{k^2}{2m}  - \frac{k}{2m} + O\left(\frac{k^3}{m^2}\right) +  O\left(\frac{1}{m}\right). \tag{2}
\end{align}

Adding Eqs. (1) and (2), we get
\begin{align}
&\log \frac{(m+k)^n}{(m+k)!} \\
&= n \log m - m \log m + m - \frac{\log(2 \pi m)}{2} - \frac{(m+n)k^2}{2m^2} + O\left(\frac{k^3 \log n}{m^2}\right) + O\left(\frac{1}{m}\right),
\end{align}
which is the expression given by Concrete Mathematics.
