Final answer:
The series ∑n=1 to [infinity] (−1)^n (ln(n))^(p/n) converges when p ≤ 1. This is because the absolute terms of the series decrease and approach zero, satisfying the conditions of the Alternating Series Test for convergence of an alternating series.
Step-by-step explanation:
The question is asking to find the values of p for which the sum ∑n=1 to [infinity] (−1)^n (ln(n))^(p/n) converges. To determine the convergence of this alternating series, we can use the Alternating Series Test (also known as Leibniz's test), which states that if the absolute terms of the series are monotonically decreasing and converge to zero, then the series converges. That is, for n sufficiently large, |ln(n)^(p/n)| should be decreasing and &lim;n→∞ ln(n)^(p/n) = 0.
Knowing that the natural logarithm function ln(x) increases slower than any positive power of x, we can deduce that ln(n) grows slower than n^(q) for any q > 0. Hence, for p > 1, (ln(n))^(p/n) will tendentially grow, and the terms won't converge to zero. For p ≤ 1, we anticipate that the terms will decrease to zero since ln(n) will not overpower the n in the denominator.
Now, let's establish that n^(1/n) approaches 1 as n goes to infinity. This means that (ln(n))^(p/n) = (ln(n))^p * (n^(1/n))^(-p) and the latter part converges to one. Therefore, the behavior of the series depends on (ln(n))^p as n approaches infinity. If p ≤ 1, (ln(n))^p approaches zero, hence the series will converge by the Alternating Series Test.In conclusion, the alternating series ∑n=1 to [infinity] (−1)^n (ln(n))^(p/n) converges when p ≤ 1, corresponding to option D).