Find the smallest value of N that will make the inequality hold for all n > N.
- $| \sqrt [ n ] { 2 n } - 1 | < 10 ^ { - 2 }$

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$| \sqrt [ n ] { 2 n } - 1 | < 10 ^ { - 2 }$$
A) 750
B) 731
C) 737
D) 733

Quizplus · Accepted Answer

We can start by simplifying the expression inside the absolute value: $\sqrt[n]{2n} - 1 = \frac{\sqrt[n]{2n} - 1}{\sqrt[n]{2n} + 1} \cdot \frac{\sqrt[n]{2n} + 1}{1} = \frac{2}{\sqrt[n]{2n} + 1}$ So our inequality becomes: $\left|\frac{2}{\sqrt[n]{2n} + 1} ight| < 10^{-2}$ Multiplying both sides by $|\sqrt[n]{2n} + 1|$: $2 < 10^{-2}|\sqrt[n]{2n} + 1|$ $200 < |\sqrt[n]{2n} + 1|$ Note that $\sqrt[n]{2n} + 1 > 1$, so we can drop the absolute value: $\sqrt[n]{2n} + 1 > 200$ $\sqrt[n]{2n} > 199$ $(2n)^{1/n} > 199$ Let $f(x)=(2^x)x^{-x}.$ Then $f'(x)=\ln(2)\cdot2^x\cdot x^{-x}-2^x\cdot(\ln x)x^{-x}-2^x\cdot x^{-x-1}.$ Thus $f'(x)=0$ implies $(\ln x)x=(x+1)\ln2.$ Note $f(x)$ increases for $0 199$ $5^n > 199^5/8^n$ Note that $5^3 = 125 < 199$, $5^4 = 625 > 199^2/8$, and $5^5 = 3125 > 199^3/8^2$. Therefore, $n = 5$ is a valid choice. Checking $n = 4$, we get: $5^4 > 199^4/8^4$ $5 > \sqrt[4]{199^4}/8$ $5 > 199/8$ which is true. Therefore, $n = 5$ is the smallest value that works, and the answer is $\boxed{ extbf{(D)}}$.

Find the Smallest Value of N That Will Make the Inequality