Find the smallest value of N that will make the inequality hold for all n > N.
- $\frac { 5 ^ { n } } { n ! } < 10 ^ { - 2 }$

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$\frac { 5 ^ { n } } { n ! } < 10 ^ { - 2 }$$
A) 16
B) 13
C) 15
D) 14

Quizplus · Accepted Answer

We can simplify the inequality using the fact that $n!>n^n$ for $n\geq 1$: $$\frac{5^n}{n!}<\frac{5^n}{n^n}$$ So, we want $\frac{5^n}{n^n}<10^{-2}$. Taking the logarithm of both sides, we get: $$n\ln(5)-n\ln(n)<-\ln(10^2)=-2\ln(10)$$ Now, we can use calculus to find the minimum of the function $f(x)=x\ln(5)-x\ln(x)$ for $x>1$. Taking the derivative and setting it equal to zero, we get: $$\ln(5)-\ln(x)-1=0$$ $$\ln\left(\frac{5}{e}\right)=\ln(x)$$ $$x=\frac{5}{e}\approx 1.84$$ This means that $f(x)$ has a minimum value at $x=\frac{5}{e}$, and since $f\left(\frac{5}{e}\right)=\frac{5}{e}\ln\left(\frac{5}{e}\right)>0$, we know that $f(x)>0$ for all $x>1$. Thus, we need: $$n\ln(5)-n\ln(n)<-2\ln(10)$$ $$n(\ln(5)-\ln(n))<-2\ln(10)$$ $$\ln\left(\frac{5}{n}\right)<-\frac{2}{\ln(10)}$$ $$\frac{5}{n}\frac{5}{0.0067}\approx 746$$ So the smallest integer value of $n$ that works is $n=747$, and the smallest integer value provided in the answer choices is $\boxed{\textbf{(C) }15}$.

Find the Smallest Value of N That Will Make the Inequality