Question

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

• A. 20
• B. 18
• C. 6
• D. 9

Answer 1

Answer: (D)

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