use the limit comparison test to determine whether the following series converges or diverges\n\nchoose the…

use the limit comparison test to determine whether the following series converges or diverges\n\nchoose the correct choice below\n\no a the series is divergent because 0 < 1/(9n² + 29) < 1/n for all n and the harmonic series is divergent\n\no b. the series is convergent because 0 < 1/(9n² + 29) < 1/n³ for all n and ∑_{n = 1}^∞ 1/n³ is convergent\n\no c the series is convergent because 0 < 1/(9n² + 29) < 1/n² for all n and ∑_{n = 1}^∞ 1/n² is convergent\n\no d. the series is divergent because 0 < 1/(9n² + 29) < 1/n² for all n and ∑_{n = 1}^∞ 1/n² is divergent

use the limit comparison test to determine whether the following series converges or diverges\n\nchoose the correct choice below\n\no a the series is divergent because 0 < 1/(9n² + 29) < 1/n for all n and the harmonic series is divergent\n\no b. the series is convergent because 0 < 1/(9n² + 29) < 1/n³ for all n and ∑_{n = 1}^∞ 1/n³ is convergent\n\no c the series is convergent because 0 < 1/(9n² + 29) < 1/n² for all n and ∑_{n = 1}^∞ 1/n² is convergent\n\no d. the series is divergent because 0 < 1/(9n² + 29) < 1/n² for all n and ∑_{n = 1}^∞ 1/n² is divergent

Answer

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.