determine if the series converges or diverges. use any method, and give a reason for your answer\n sum _ { n…

determine if the series converges or diverges. use any method, and give a reason for your answer\n sum _ { n = 1 } ^ { infty } \frac { 2 n } { 9 n - 2 } \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice\na. the series diverges because the limit found in the nth - term test is (simplify your answer.)\nb. because ( \frac { 2 n } { 9 n - 2 } geq n ) and ( sum _ { n = 1 } ^ { infty } n ) diverges, the series diverges by the direct comparison test\nc. the series converges because the limit found in the nth - term test is (simplify your answer.)\nd. because ( \frac { 2 n } { 9 n - 2 } leq \frac { 1 } { n } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n } ) converges, the series converges by the direct comparison test

determine if the series converges or diverges. use any method, and give a reason for your answer\n sum _ { n = 1 } ^ { infty } \frac { 2 n } { 9 n - 2 } \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice\na. the series diverges because the limit found in the nth - term test is (simplify your answer.)\nb. because ( \frac { 2 n } { 9 n - 2 } geq n ) and ( sum _ { n = 1 } ^ { infty } n ) diverges, the series diverges by the direct comparison test\nc. the series converges because the limit found in the nth - term test is (simplify your answer.)\nd. because ( \frac { 2 n } { 9 n - 2 } leq \frac { 1 } { n } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n } ) converges, the series converges by the direct comparison test

Answer

Explanation:

Step1: Apply the nth - Term Test

The nth - Term Test for divergence states that if (\lim_{n\rightarrow\infty}a_{n}\neq0), then the series (\sum_{n = 1}^{\infty}a_{n}) diverges. For the series (\sum_{n=1}^{\infty}\frac{2n}{9n - 2}), we find the limit (\lim_{n\rightarrow\infty}\frac{2n}{9n - 2}).

Step2: Simplify the limit

Divide both the numerator and denominator by (n): (\lim_{n\rightarrow\infty}\frac{2n/n}{(9n - 2)/n}=\lim_{n\rightarrow\infty}\frac{2}{9-\frac{2}{n}}). As (n\rightarrow\infty), (\lim_{n\rightarrow\infty}\frac{2}{n}=0). So, (\lim_{n\rightarrow\infty}\frac{2}{9-\frac{2}{n}}=\frac{2}{9}\neq0).

Answer:

A. The series diverges because the limit found in the nth - Term Test is (\frac{2}{9})