Question

use the nth - term test for divergence to show that the series is divergent, or state that the test is inconclusive.

sum _ { n = 1 } ^ { infty } cos \frac { 12 } { n }

choose the correct answer below.

a. the test is inconclusive because ( lim _ { n
ightarrow infty } cos \frac { 12 } { n } = )

b. the series diverges because ( lim _ { n
ightarrow infty } cos \frac { 12 } { n } ) exists and is equal to

c. the series diverges because ( lim _ { n
ightarrow infty } cos \frac { 12 } { n } = - infty ) and fails to exist

d. the series diverges because ( lim _ { n
ightarrow infty } cos \frac { 12 } { n } = infty ) and fails to exist

Explanation:

Step1: Find the limit of the general term

We know that \(\lim_{n
ightarrow\infty}\frac{12}{n} = 0\).
By the continuity of the cosine function \(y = \cos x\), \(\lim_{n
ightarrow\infty}\cos\frac{12}{n}=\cos(\lim_{n
ightarrow\infty}\frac{12}{n})\).
Since \(\lim_{n
ightarrow\infty}\frac{12}{n} = 0\), then \(\lim_{n
ightarrow\infty}\cos\frac{12}{n}=\cos(0)\).

Step2: Calculate the value of \(\cos(0)\)

We know that \(\cos(0)=1
eq0\).
According to the nth - term test for divergence: If \(\sum_{n = 1}^{\infty}a_{n}\) is a series and \(\lim_{n
ightarrow\infty}a_{n}
eq0\), then the series \(\sum_{n = 1}^{\infty}a_{n}\) diverges. Here \(a_{n}=\cos\frac{12}{n}\).

Answer:

B. The series diverges because \(\lim_{n
ightarrow\infty}\cos\frac{12}{n}\) exists and is equal to \(1\)