The question is as following:
For what values of $p$ is the series convergent? $$ \sum_{n=2}^\infty (-1)^{n-1}\frac{(\ln n)^p}{n^2}$$
What I have done so far:
Take Absolute Value (Test for Absolute Convergent)
Using Integration By Parts repeatedly (Do Integral Test) $$\int_{2}^\infty\frac{(\ln x)^p}{x^2}dx= \lim_{t\to\infty}\left[(\ln x)^p(-\frac{1}{x})+p(\ln x)^{p-1}(- \frac{1}{x})+p(p-1)(\ln x)^{p-2}(- \frac{1}{x})+...+p!(- \frac{1}{x})\right]_{2}^{t}$$
I found that the value of $p$ seems have no effect on the convergence of the series and my answer for this question is $[0,\infty)$ but when I tested it with WolframAlpha, I found it when $p=n$, it is not convergent.
And thus I know I may have made some mistakes.
So can anyone point out which part did I do it wrongly? It would be really appreciated :D
Thanks for you help!
It is always convergent by comparison, indeed $$\frac{(\ln n)^p}{n^2} = o \left(\frac 1{n^{\frac32}}\right)$$