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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!

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What book would that be? – Will Jagy Apr 14 '12 at 1:09
It's unreleased as of yet, a book on percolation theory - I have a photocopy of a small outtake as it's not yet finished (perhaps I should have said "a chapter of a book"). Is the claim incorrect? – Spyam Apr 14 '12 at 3:00
up vote 1 down vote accepted

The LHS is $a_k=\frac1{2k}\cdot k!\cdot \mathrm e^{k}\cdot \mathrm P(X_k\leqslant k-3)$ where $X_k$ is a random variable with Poisson distribution of parameter $k$.

Since $X_k$ has the distribution of the sum of $k$ i.i.d. random variables with Poisson distribution of parameter $1$, the classical central limit theorem yields $\mathrm P(X_k\leqslant k-3)\to\frac12$.

Thus, $a_k\sim\frac1{4k}\cdot k!\cdot \mathrm e^{k}$. Stirling's formula $k!\cdot\mathrm e^{k}\sim\sqrt{2\pi k}\cdot k^k$ yields the equivalent in the RHS.

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