Question

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

Explanation:

Step1: Apply the nth - Term Test

The nth - Term Test for divergence states that if \(\lim_{n
ightarrow\infty}a_{n}
eq0\), 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 \(\lim_{n
ightarrow\infty}\frac{2n}{9n - 2}\).

Step2: Simplify the limit

Divide both the numerator and denominator by \(n\): \(\lim_{n
ightarrow\infty}\frac{2n/n}{(9n - 2)/n}=\lim_{n
ightarrow\infty}\frac{2}{9-\frac{2}{n}}\).
As \(n
ightarrow\infty\), \(\lim_{n
ightarrow\infty}\frac{2}{n}=0\). So, \(\lim_{n
ightarrow\infty}\frac{2}{9-\frac{2}{n}}=\frac{2}{9}
eq0\).

Answer:

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