Consider $$ e^{N/d}\frac{d}{2N}\sum_{n=0}^{N-1}\left(\frac{d}{N}\right)^nn! $$ I know its asyptotics when $d<e$ but I haven't suceeded for $d\geq e$. Can you do better and work out its big $N$ asymptotics for $d\geq e$?
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What exactly is your problem? – Alex Oct 01 '16 at 11:52
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@Alex I cannot get the big $N$asymptotics for $d\geq e$. – PhoenixPerson Oct 01 '16 at 12:03
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Have you tried exploiting $$\sum_{n=0}^{N-1}\alpha^n n! = \int_{0}^{+\infty}e^{-x}\sum_{n=0}^{N-1}(\alpha x)^n = \int_{0}^{+\infty}\frac{(\alpha x)^N-1}{\alpha x-1}e^{-x},dx=\frac{1}{\alpha}\int_{0}^{+\infty}\frac{z^N-1}{z-1}e^{-z/\alpha},dz$$ and Laplace method? – Jack D'Aurizio Oct 01 '16 at 13:20
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@JackD'Aurizio I have precisely come from the last equation on the right. I am not very familiar with Laplace and saddles unfortunately... – PhoenixPerson Oct 01 '16 at 13:57
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Then try this simple approach: on the interval $(0,1)$ we have $e^{-z/\alpha}\leq 1$, hence such interval contributes to the global integral by at most $H_N\sim\log(N)$. On the interval $(1,+\infty)$, the integral is very close to $\int_{1}^{+\infty}z^{N-1}e^{-z/\alpha},dz$ and $$\int_{0}^{+\infty}z^{N-1}e^{-z/\alpha},dz$$ is easy to compute through the $\Gamma$ function. – Jack D'Aurizio Oct 01 '16 at 14:11