Comparing powers and factorials

Question

Let $a_n = \frac {1000^n}{n!}$ for $n \in N$. Then $a_n$ is greatest for:-

(A) $n=998$ (B) $n=999$ (C) $n=1000$ (D) $n=1001$

How to approach? Need hints.

Answer 1

In general the maximum value is between $n$ and $n-1$.

$sup(\frac{c^n}{n!})$ and using Stirling approximation $n! \approx n^n e^{-n} \sqrt{2 \pi n}$.

Which is approximately $sup(\frac{c^n}{n^n e^{-n} \sqrt{2 \pi n}})$ , the maximum value is when $(\frac{c^n}{n^n e^{-n} \sqrt{2 \pi n}})' =0$, we reach at $\frac{e^n n^{-n-\frac{1}{2}} c^n}{\sqrt{2 \pi }}+\frac{e^n n^{-n-\frac{1}{2}} c^n \log (c)}{\sqrt{2 \pi }}+\frac{e^n n^{-n-\frac{1}{2}} c^n \left(\frac{-n-\frac{1}{2}}{n}-\log (n)\right)}{\sqrt{2 \pi }}=0$ with a little bit of manipulation we arrive at $\log (c)-\frac{1}{2 n}-\log (n)=0$,which gives the answer $n=-\frac{1}{2 W\left(-\frac{1}{2 c}\right)}$, which can be easily bounded between $n$ and $n-1$.