Use the Root Test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \left( \frac { 4 n ^ { 1 / n } - 1 } { 3 n ^ { 1 / n } - 1 } \right) ^ { n }$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

Using the root test, we have:
$$\lim_{n\to\infty} \sqrt[n]{\left(\frac{4n^{1/n}-1}{3n^{1/n}-1}\right)^n}=\lim_{n\to\infty}\frac{4n^{1/n}-1}{3n^{1/n}-1}$$
Dividing the numerator and denominator by $n^{1/n}$, we get:
$$\lim_{n\to\infty}\frac{4-\frac{1}{n^{1/n}}}{3-\frac{1}{n^{1/n}}}=\frac{4}{3}$$
Since the limit is greater than 1, by the root test, the series diverges. Therefore, the answer is (A) Diverges.

Use the Root Test to Determine If the Series Converges ∑n=1∞(4n1/n−13n1/n−1)n\sum _ { n = 1 } ^ { \infty } \left( \frac { 4 n ^ { 1 / n } - 1 } { 3 n ^ { 1 / n } - 1 } \right) ^ { n }n=1∑∞​(3n1/n−14n1/n−1​)n

Use the Root Test to Determine If the Series Converges $\sum _ { n = 1 } ^ { \infty } \left( \frac { 4 n ^ { 1 / n } - 1 } { 3 n ^ { 1 / n } - 1 } \right) ^ { n }$