choose the correct answer below and, if necessary, fill in the answer box to complete your choice.\na. the…

choose the correct answer below and, if necessary, fill in the answer box to complete your choice.\na. the series converges absolutely because the corresponding series of absolute values is (comparative with )\n\nb. the series converges conditionally per the alternating series test and the comparison test with \n\nc. the series converges conditionally per alternating series test and because the limit used in the nth-term test is \n\nd. the series diverges because the limit used in the ratio test is not less than or equal to 1\n\ne. the series diverges because the limit used in the nth-term test is not zero\n\nf. the series converges absolutely because the corresponding series of absolute values is a p - series with p =\n\n

choose the correct answer below and, if necessary, fill in the answer box to complete your choice.\na. the series converges absolutely because the corresponding series of absolute values is (comparative with )\n\nb. the series converges conditionally per the alternating series test and the comparison test with \n\nc. the series converges conditionally per alternating series test and because the limit used in the nth-term test is \n\nd. the series diverges because the limit used in the ratio test is not less than or equal to 1\n\ne. the series diverges because the limit used in the nth-term test is not zero\n\nf. the series converges absolutely because the corresponding series of absolute values is a p - series with p =\n\n

Answer

Explanation:

Step1: Analyze the series of absolute values

The series of absolute values is (\sum_{n = 1}^{\infty}\frac{1}{n}). This is a (p -)series with (p=1).

Step2: Recall the (p -)series convergence criterion

For a (p -)series (\sum_{n = 1}^{\infty}\frac{1}{n^{p}}), it converges if (p>1) and diverges if (p\leq1). Since (p = 1) for (\sum_{n = 1}^{\infty}\frac{1}{n}), the series of absolute values diverges.

Step3: Apply the Alternating Series Test

For the alternating series (\sum_{n = 1}^{\infty}\frac{(- 1)^{n + 1}}{n}), let (a_{n}=\frac{1}{n}).

  • Condition 1: (\lim_{n\rightarrow\infty}a_{n}) (\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\frac{1}{n}=0)
  • Condition 2: (a_{n+1}\leq a_{n}) for all (n) sufficiently large (a_{n}=\frac{1}{n}), (a_{n + 1}=\frac{1}{n+1}), and (\frac{1}{n+1}<\frac{1}{n}) for (n\geq1)

Since the series of absolute values (\sum_{n = 1}^{\infty}\left|\frac{(-1)^{n + 1}}{n}\right|=\sum_{n = 1}^{\infty}\frac{1}{n}) diverges (by the (p -)series test with (p = 1)) and the alternating series (\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}) converges (by the Alternating Series Test), the original series converges conditionally.

Answer:

C. The series converges conditionally per the Alternating Series Test and because the limit used in the (n)th - Term Test is (0)