For what values of $p$ does the series
$$\sum_{n=2}^{\infty}{(-1)}^n \frac{ {(\ln n)}^p }{n}$$
converge?
Using the alternating series test the only way I made the it diverge was when I replaced $p$ with $n$, so is there no number that makes the series diverge?
Do all numbers make the series converge?
I used desmos 
and set up a slider for $p$ and tried to make the limit as $n$ goes to infinity do anything other than go to $0$ (alternating series test says if it goes to $0$ then it converges), but if I scrolled out enough it always went to $0$.
So my question is: $(1)$ Is there a value of p that makes the series diverge (limit disregarding the $-1$ go to anything other than $0$) or do all values of $p$ make this go to $0$ and therefore the series always converges? $(2)$ How would you figure out what that number is (assuming it exists)?
Best Answer
For any fixed $p$, the sequence $\frac{(\ln n)^p}n$ is eventually decreasing, and so by Leibniz, the series $\sum_n(-1)^n\frac{(\ln n)^p}n$ converges.