Tom K. answered 11/18/20
Knowledgeable and Friendly Math and Statistics Tutor
The first series converges. You can use the integral test on 1/n^3
The third series diverges. The sequence does not converge to 0.
The fourth series converges. The absolute value is bounded above by 1/n!, which has a ratio which converges to 0 (in fact, its sum is e).
Thus, let's look at the second function.
As we have (-1)^n, this is an alternating series.
As lim n -> ∞ ln(n) = ∞, lim n -> ∞ 1/ln(n) = 0
ln(n) < n, so 1/ln(n) > 1/n, whose sum diverges. (You can see this from the integral test, the integral of 1/n is ln(n) which again has a limit of ∞).
Thus, this series has conditional convergence.