Question

use the limit comparison test to determine whether the following series converges or diverges

choose the correct choice below

o a the series is divergent because 0 < 1/(9n² + 29) < 1/n for all n and the harmonic series is divergent

o b. the series is convergent because 0 < 1/(9n² + 29) < 1/n³ for all n and ∑_{n = 1}^∞ 1/n³ is convergent

o c the series is convergent because 0 < 1/(9n² + 29) < 1/n² for all n and ∑_{n = 1}^∞ 1/n² is convergent

o d. the series is divergent because 0 < 1/(9n² + 29) < 1/n² for all n and ∑_{n = 1}^∞ 1/n² is divergent

Explanation:

Step1: Recall the comparison test for series

For two series \(\sum_{n = 1}^{\infty}a_{n}\) and \(\sum_{n=1}^{\infty}b_{n}\) with \(0\leq a_{n}\leq b_{n}\) for all \(n\):

• If \(\sum_{n = 1}^{\infty}b_{n}\) converges, then \(\sum_{n=1}^{\infty}a_{n}\) converges.
• If \(\sum_{n = 1}^{\infty}a_{n}\) diverges, then \(\sum_{n=1}^{\infty}b_{n}\) diverges.

Step2: Analyze the given series \(\sum_{n = 1}^{\infty}\frac{1}{9n^{2}+29}\)

We know that \(9n^{2}+29>9n^{2}\), so \(\frac{1}{9n^{2}+29}<\frac{1}{9n^{2}}\).

Step3: Consider the \(p -\)series \(\sum_{n = 1}^{\infty}\frac{1}{n^{p}}\)

The \(p -\)series \(\sum_{n = 1}^{\infty}\frac{1}{n^{p}}\) converges if \(p> 1\) and diverges if \(p\leq1\). For the series \(\sum_{n = 1}^{\infty}\frac{1}{9n^{2}}=\frac{1}{9}\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\), since \(p = 2>1\), \(\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\) converges. Then \(\frac{1}{9}\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\) also converges.

Step4: Apply the comparison test

Since \(0<\frac{1}{9n^{2}+29}<\frac{1}{9n^{2}}\) and \(\sum_{n = 1}^{\infty}\frac{1}{9n^{2}}\) converges, by the comparison test, \(\sum_{n = 1}^{\infty}\frac{1}{9n^{2}+29}\) converges.

Answer:

B. The series is convergent because \(0\leq\frac{1}{9n^{2}+29}\leq\frac{1}{n^{2}}\) for all \(n\) and \(\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\) is convergent.