A book I'm reading claims that $\frac{1}{2}(k-1)!\sum \limits_{j=0}^{k-3} \frac{k^j}{j!} \sim (\pi / 8)^{1/2}k^{k-\frac{1}{2}}$ as $k \to \infty$. I can get most of the expression to work out nicely but I can't convince myself that the limit for the sum is actually $e^k$, the "obvious" limit, because as the partial sum for the Taylor series becomes larger, i.e. as $k \to \infty$, the function we're approximating $(e^k)$ also becomes larger, so sure, we might have more terms in the sum but we're also approximating something larger and larger so the overall error might not actually get smaller fast enough for convergence to $e^k$. I'm certain it does, but I just want to see why. So this is what I have so far (I think I've also lost a factor of $2$ somewhere), basically doing nothing but applying Stirling's approximation and the alternative definition for $e^x$ as a limit of $(1+x/n)^n$ as $n \to \infty$:
$\frac{1}{2}(k-1)!\sum \limits_{j=0}^{k-3} \frac{k^j}{j!} \sim \frac{1}{2}\sqrt{2\pi (k-1)}\left(\frac{k-1}{e}\right)^{k-1} \sum_j \sim (\pi k/2)^{1/2} e^{1-k} k^{k-1} \left(1-\frac{1}{k}\right)^{-1} \left(1-\frac{1}{k}\right)^{k} \sum_j \sim (\pi/2)^{1/2} k^{k-1/2} e^{-1} e^{1-k} \sum_j \sim (\pi/2)^{1/2} k^{k-1/2} e^{-k} \sum_j$
What's the next step? And where did I lose that $2$? I know I could add an extra factor of $k$ to the initial factorial and then make the Stirling's approximation part slightly tidier, but I don't think that makes any difference. Thanks for your help!