# How to find the interval of p such that the series is convergent?

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:

1. Take Absolute Value (Test for Absolute Convergent)

2. 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!

• $p$ is supposed to be a fixed real number. You can't set it to $n$. – saulspatz May 4 '18 at 15:13
• um...so is it my answer is correct? @saulspatz – Cluyeia May 4 '18 at 15:17
• It converges for $p\ge0$ by the alternating series test. What about negative $p$? – saulspatz May 4 '18 at 15:20
• oh that's also convergent by the alternating series test. thanks!! :) – Cluyeia May 4 '18 at 15:24

It is always convergent by comparison, indeed $$\frac{(\ln n)^p}{n^2} = o \left(\frac 1{n^{\frac32}}\right)$$