If $\displaystyle a_{n} = \frac{1000^n}{n!}$ for $n\in \mathbb{N}\;,$ Then $a_{n}$ is greatest when

$\bf{Options::}\;\;(a)\;\; n=997\;\;\;\; (b)\;\; n=998\;\;\;\; (c)\;\; n=999\;\;\;\;(a)\;\; n=1000\;\;\;\;$

$\bf{My\; Try::}$ 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 $n=1000$

  • $\begingroup$ (a) $n=1000$ may be a typo. $\endgroup$ – choco_addicted Feb 17 '16 at 15:49

What you have concluded is that:


For $n=997,998$. This is concluded from your equation (just algebra).

So plug in $n=998$

To get:


Then if you plug in $n=999$ you will see that:



For $n=999$, $\dfrac{a_{1000}}{a_{999}}=1$, hence $a_{999}=a_{1000}$, while $\dfrac{a_{999}}{a_{998}}>1$, as you established it, hence $a_{999}>a_{998}$.


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