Find the smallest value of N that will make the inequality hold for all n > N.
- $\frac { 10 ^ { n } } { n ! } < 10 ^ { - 3 }$

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$\frac { 10 ^ { n } } { n ! } < 10 ^ { - 3 }$$
A) 21
B) 31
C) 20
D) 30

Quizplus · Accepted Answer

We can simplify the inequality by taking the logarithm of both sides: $\log{\left(\dfrac{10^n}{n!}\right)}<\log{\left(10^{-3}\right)}$ $\Rightarrow n\log{10}-\log{(n!)}<-3$ $\Rightarrow n-1- \sum\limits_{k=2}^{n}\log{k}<-3$ (using the approximation $\log{n!}\approx n\log{n}-n$) $\Rightarrow \sum\limits_{k=2}^{n}\log{k}>n+2$ We can now estimate the left-hand side of the inequality using the fact that the sum is the area under the curve $y=\log{x}$ from $x=2$ to $x=n$. This area can be approximated by the area of the rectangle with height $\log{2}$ and base $n-2$, which gives: $\sum\limits_{k=2}^{n}\log{k}>\log{2}(n-2)$ Substituting this approximation into our inequality, we get: $\log{2}(n-2)>n+2$ $\Rightarrow n>\dfrac{2+\log{2}}{\log{2}-1}\approx 30.7$ Since we need to find the smallest integer value of $n$ that satisfies the inequality, the answer is $n=31$.

Find the Smallest Value of N That Will Make the Inequality