Provide an appropriate response.
-It can be shown that $\sqrt [ n ] { n ! } \approx \frac { n } { e }$ for large values of $n$ . Find the smallest value of $N$ such that $\frac { \sqrt [ n ] { n ! } } { \frac { n } { e } } - 1 < 10 ^ { - 1 }$ for all $\mathrm { n } > \mathrm { N }$ .

Question

Provide an appropriate response.
-It can be shown that $\sqrt [ n ] { n ! } \approx \frac { n } { e }$ for large values of $n$. Find the smallest value of $N$ such that $\frac { \sqrt [ n ] { n ! } } { \frac { n } { e } } - 1 < 10 ^ { - 1 }$ for all $\mathrm { n } > \mathrm { N }$.
A) 29
B) 33
C) 27
D) 22

Quizplus · Accepted Answer

First, we can simplify the given expression as:
$$\frac{\sqrt[n]{n!}}{\frac{n}{e}}-1 = \frac{\sqrt[n]{n!}-\frac{n}{e}}{\frac{n}{e}}$$
Using Stirling's approximation, we have:
$$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$
Taking the nth root and simplifying, we get:
$$\sqrt[n]{n!} \sim \frac{n}{e}\sqrt[n]{\frac{2\pi}{n}}$$
Substituting this in our expression and simplifying, we get:
$$\frac{\sqrt[n]{n!}-\frac{n}{e}}{\frac{n}{e}} \sim \frac{\frac{n}{e}\sqrt[n]{\frac{2\pi}{n}}-\frac{n}{e}}{\frac{n}{e}} = \sqrt[n]{\frac{2\pi}{n}}-1$$
For the given expression to be less than $10^{-1}$, we need $\sqrt[n]{\frac{2\pi}{n}}-1<10^{-1}$. 
Solving this inequality, we get $n>27.2$, so the smallest integer value of $n$ satisfying the inequality is $n=28$. Therefore, the smallest value of $N$ for which the inequality holds for all $n>N$ is $N=27$. Thus, the answer is (C) 27.

Provide an Appropriate Response n!n≈ne\sqrt [ n ] { n ! } \approx \frac { n } { e }nn!​≈en​

Provide an Appropriate Response $\sqrt [ n ] { n ! } \approx \frac { n } { e }$