$\text { Determine whether the series } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \ln 2 n } \text { converges conditionally or }$ absolutely, or diverges.
A)The series converges absolutely.
B)The series diverges.
C)The series converges absolutely but does not converge conditionally.
D)The series converges conditionally but does not converge absolutely.

Question

Quizplus · Accepted Answer

We can use the alternating series test to show that the series converges:
The terms approach 0 (since $\ln(2n) > 1$ for $n\ge2$ ) and $(1/\ln(2n))$ is decreasing as n increases.
To see why the series converges conditionally, let $a_n=\frac{1}{\ln(2n)}$ and $b_n=\frac{1}{\ln(n)}.$
Then, $\lim_{n	o\infty}\frac{a_n}{b_n}=\lim_{n	o\infty}\frac{\ln(n)}{\ln(2n)}=\lim_{n	o\infty}\frac{\ln n}{\ln2+\ln n} = 1/(\ln 2)$.
Since the harmonic series diverges diverges, we have by the limit comparison test that our series also diverges when $b_n$ is used as the reference sequence.
So the series converges conditionally, but not absolutely.