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determine if the series converges or diverges. give a reason for your a…

Question

determine if the series converges or diverges. give a reason for your answer.

$$ sum _ { n = 1 } ^ { infty } \frac { 2 n } { 3 n - 8 } $$

select the correct choice below and, if necessary, fill in the answer box to complete your choice

a. the direct comparison test with $$ sum n $$ shows that the series diverges

b. the series converges because the limit found in the nth-term test is (type an integer or a fraction.)

c. the series diverges because the limit found in the nth-term test is (type an integer or a fraction )

d. the direct comparison test with $$ sum \frac { 1 } { n } $$ shows that the series converges

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 \(\sum_{n=1}^{\infty}\frac{2n}{3n - 8}\), we find the limit \(\lim_{n
ightarrow\infty}\frac{2n}{3n - 8}\).
Divide both the numerator and denominator by \(n\):

$$ LATEXBLOCK0 $$

Step2: Evaluate the limit

As \(n
ightarrow\infty\), \(\lim_{n
ightarrow\infty}\frac{8}{n}=0\).
Using the limit rule \(\lim_{n
ightarrow\infty}(a - b)=\lim_{n
ightarrow\infty}a-\lim_{n
ightarrow\infty}b\) (where \(a = 3\) and \(b=\frac{8}{n}\)), we have \(\lim_{n
ightarrow\infty}\frac{2}{3-\frac{8}{n}}=\frac{2}{3-0}=\frac{2}{3}\)

Answer:

C. The series diverges because the limit found in the nth - Term Test is \(\frac{2}{3}\)