Let $a_n=\frac{10^n}{n !}$ for $n=1,2,3, \ldots$ then the greatest value of $n$ for which $a_n$ is the…
Let $a_n=\frac{10^n}{n !}$ for $n=1,2,3, \ldots$ then the greatest value of $n$ for which $a_n$ is the greatest is
11
20
10
8
Solution
Given, $a_n=\frac{10^n}{n !}, n=1,2,3, \ldots$, here we see that when we increase the value of $n$ like as $1,2,3, \ldots$ the value of $a_n$ increases but when we reach at $n=9$ or 10 the value of $a_n$ remain unchanged, $i e$, minor difference in after decimal places and when we cross the value $n=10 \mathrm{ie}$, $n=11$, then we see that the value of $a_n$ is monotonically decreasing.
Hence, $a_n$ have its maximum value at $n=9$ or 10.