By calculating an appropriate number of terms, determine if the series converges or diverges. If it converges, find the limit $L$ and the smallest integer $N$ such that $\left| a _ { n } - L \right| < 0.01$ for $n \geq N$ ; otherwise indicate divergence.
- $a _ { n } = 1539 ^ { 1 / n }$

Question

By calculating an appropriate number of terms, determine if the series converges or diverges. If it converges, find the limit $L$ and the smallest integer $N$ such that $\left| a _ { n } - L \right| < 0.01$ for $n \geq N$; otherwise indicate divergence.
-$$a _ { n } = 1539 ^ { 1 / n }$$
A) L = 1, N = 738
B) L = 1, N = 1549
C) L = 0, N = 1549
D) diverges

Quizplus · Accepted Answer

Taking the limit as $n$ approaches infinity, we have:

$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}1539^{1/n}=1539^{0}=1$$

Therefore, the series converges to 1.

To find the smallest integer $N$ such that $\left| a _ { n } - 1 \right| < 0.01$ for $n \geq N$, we can solve the inequality:

$$\left| a _ { n } - 1 \right| < 0.01$$

$$-0.01 < a_n -1 < 0.01$$

$$0.99 < a_n < 1.01$$

$$1539^{1/n} < 1.01$$

Taking natural logarithm on both sides:

$$\frac{1}{n}\ln(1539) < \ln(1.01)$$

$$n > \frac{\ln(1539)}{\ln(1.01)} \approx 738.3$$

Therefore, the smallest integer $N$ is 739 or greater. The best choice is A) L=1, N=738.

By Calculating an Appropriate Number of Terms, Determine If the Series