Does the series Sigma(n=1 to infinity) (-1)^n ln[(3n+1)/(3n-2)] converge?

Answer 1

The alternating series test says that the series sum I= 1 to I infinity of (-1)^i (b_i) where (b_i) >= 0 converges if (b_i) is decreasing and converges to 0.

in this example (3n+1)/3n-2) is decreasing and converges to 1. Since ln is an increasing function, ln((3n+1)/((3n-2)) is also decreasing and converges to ln1=0. So by Alternating Series test this series converges.

Answer 2

(3n+1)/(3n-2) converges to 3n/3n = 1

ln1 = 0

(-1)^n time ln1 = (-1)^n times zero = zero

the wholle expression converges to zero as n approaches infinity

You might be tempted to say it diverges due to the (-1)^n term as it oscilates between 1 and -1

but that's misleading as a finite number times zero is zero