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> If $\displaystyle a\_{n} = \frac{1000^n}{n!}$ for $n\in \mathbb{N}\;,$ Then $a\_{n}$ is greatest when
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> $\bf{Options::}\;\;(a)\;\; n=997\;\;\;\; (b)\;\; n=998\;\;\;\; (c)\;\; n=999\;\;\;\;(a)\;\; n=1000\;\;\;\;$
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$\bf{My\; Try::}$ Given $\displaystyle a\_{n} = \frac{1000^n}{n!}\;,$ Then $\displaystyle a\_{n+1} = \frac{1000^{n+1}}{(n+1)!}\;,$

So $\displaystyle\frac{a\_{n+1}}{a\_{n}} = \frac{1000}{n+1}>1$ for $n=997,998$

So $a\_{n}$ is $\bf{\max}$ when $n=998$

But answer given is $a\_{n}$ is $\max$ when $n=999$ and $n=1000$