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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!

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    $\begingroup$ $p$ is supposed to be a fixed real number. You can't set it to $n$. $\endgroup$ – saulspatz May 4 '18 at 15:13
  • $\begingroup$ um...so is it my answer is correct? @saulspatz $\endgroup$ – Cluyeia May 4 '18 at 15:17
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    $\begingroup$ It converges for $p\ge0$ by the alternating series test. What about negative $p$? $\endgroup$ – saulspatz May 4 '18 at 15:20
  • $\begingroup$ oh that's also convergent by the alternating series test. thanks!! :) $\endgroup$ – Cluyeia May 4 '18 at 15:24
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It is always convergent by comparison, indeed $$\frac{(\ln n)^p}{n^2} = o \left(\frac 1{n^{\frac32}}\right)$$

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  • $\begingroup$ umm may I know what comparison have you made? I'm only a Yr1 student and don't know too much about the little o notation... $\endgroup$ – Cluyeia May 4 '18 at 15:19
  • $\begingroup$ Direct test comparison series en.wikipedia.org/wiki/Direct_comparison_test $\endgroup$ – Youem May 4 '18 at 15:22

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