Examine the two series below for absolute convergence (A), convergence that is not absolute (C), or divergence (D).
1) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { \ln ( n + 1 ) }$
2) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { ( \ln ( n + 1 ) ) ^ { 2 } }$
A)1A, 2A
B)1A, 2C
C)1A, 2D
D)1C, 2A
E)1C, 2C
F)1C, 2D
G)1D, 2A
H)1D, 2C

Question

Examine the two series below for absolute convergence (A), convergence that is not absolute (C), or divergence (D).
1) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { \ln ( n + 1 ) }$ 
2) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { ( \ln ( n + 1 ) ) ^ { 2 } }$ 
A)1A, 2A
B)1A, 2C
C)1A, 2D
D)1C, 2A
E)1C, 2C
F)1C, 2D
G)1D, 2A
H)1D, 2C

Quizplus · Accepted Answer

1) The series $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { \ln ( n + 1 ) }$ is an alternating series. The terms $\frac { 1 } { \ln ( n + 1 ) }$ decrease to 0 as $n$ increases, so the series converges by the Alternating Series Test. However, it does not converge absolutely because the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { \ln ( n + 1 ) }$ diverges by comparison with the harmonic series or by the integral test.2) The series $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { ( \ln ( n + 1 ) ) ^ { 2 } }$ also converges by the Alternating Series Test since the terms $\frac { 1 } { ( \ln ( n + 1 ) ) ^ { 2 } }$ decrease to 0. However, it does not converge absolutely for similar reasons as the first series, but the divergence is more evident due to the squared denominator, which makes the comparison to the p-series (with p=2) or the integral test show divergence more clearly.

Examine the Two Series Below for Absolute Convergence (A), Convergence ∑n=1∞(−1)n−1ln⁡(n+1)\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { \ln ( n + 1 ) }∑n=1∞​ln(n+1)(−1)n−1​

Examine the Two Series Below for Absolute Convergence (A), Convergence $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { \ln ( n + 1 ) }$