Question

Let $a_n=10^n / n$ ! for $n \geq 1$. Then $a_n$ take the greatest value when $n$ equals

• A. 20
• B. 18
• C. 6
• D. 9

Answer 1

Answer: (D)

We have $\frac{a_n}{a_{n+1}}=\frac{10^n}{n !} \times \frac{(n+1) !}{10^{n+1}}=\frac{n+1}{10}$.
$\text { Note that } \frac{a_1}{a_2}<1, \frac{a_2}{a_3}<1, \frac{a_3}{a_4}<1, \frac{a_4}{a_5}<1, $
$\ldots \frac{a_8}{a_9}<1, \frac{a_9}{a_{10}}=1, \frac{a_{10}}{a_{11}}>1, \frac{a_{11}}{a_{12}}>1, \ldots$
Thus, $ a_1< a_2< a_3<\ldots< a_9=a_{10} >a_{11} >a_{12}$
$\therefore a_n$ is greatest if $n=9$ or 10 .