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

We begin by using the given approximation and taking the natural logarithm of both sides: \begin{align*} \ln(\sqrt[n]{n!}) &\approx \ln\left(\frac{n}{e}\right) \ \frac{1}{n} \ln(n!) &\approx \ln(n) - \ln(e) \ \frac{1}{n} \sum_{i=1}^n \ln(i) &\approx \ln(n) - 1 \end{align*} Now, we use the fact that $\ln(x+1) \leq x$ for all $x>-1$ to obtain an upper bound: \begin{align*} \frac{1}{n} \sum_{i=1}^n \ln(i) &\leq \frac{1}{n} \sum_{i=1}^n \ln(i+1) \ &= \frac{1}{n} \left(\ln(2) + \ln(3) + \cdots + \ln(n+1)\right) \ &= \frac{\ln(n+1)}{n} - \frac{\ln(1)}{n} \ &= \frac{\ln(n+1)}{n} \ &\approx \frac{\ln(n)}{n} \ &< \frac{1}{n} \quad \text{(for } n > e \text{)} \end{align*} We want to find the smallest $N$ such that \begin{align*} \frac{\sqrt[n]{n!}}{\frac{n}{e}} - 1 &< 10^{-1} \ \sqrt[n]{n!} &< \frac{1.1n}{e} \end{align*} Taking logarithms again and using the upper bound derived above, we have \begin{align*} \ln(\sqrt[n]{n!}) &< \ln\left(\frac{1.1n}{e}\right) \ \frac{1}{n} \sum_{i=1}^n \ln(i) &< \ln(1.1n) - 1 \end{align*} For $n \geq 27$, we have $\ln(1.1n) - 1 < \frac{1}{n}$, so $$\frac{\sqrt[n]{n!}}{\frac{n}{e}} - 1 < \frac{\ln(n)}{n} < \frac{1}{n} < 0.1$$ thus $N$ must be at most $26$. Checking the remaining answer choices, we see that $\boxed{\textbf{(C) }27}$ is the smallest value of $N$ that satisfies the conditions.

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

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