$\sum _{ n=10 }^{ \infty }{ (-1)^ n } \frac { n^ 2 }{ ln(n) } $

How to find whether it diverges or converges

My attempt

By using alternating series test:

$\displaystyle \lim_{n \rightarrow \infty}$ $\frac { n^ 2 }{ ln(n) } $ = $\infty$.. It is not equal to zero... so the series diverges.


I don't get the reference to the alternating series test; since $(-1)^n\frac{n^2}{\log{n}}\not\to 0$, the series diverges.

  • $\begingroup$ But, I need to prove using alternating series test... 1) proving using limits and 2) proving using direct comparison test $\endgroup$ – tien lee Apr 29 '18 at 16:03
  • $\begingroup$ @tienlee If the terms of the series do not approach zero, then the series diverges $\endgroup$ – Mark Viola Apr 29 '18 at 16:04
  • $\begingroup$ You don't need any alternating series test to prove this, it's straight forward calculation $\endgroup$ – JustDroppedIn Apr 29 '18 at 16:05
  • $\begingroup$ But, my professor told to show the alternating series test when ever there is $(-1)^n$ and I need to show this during the exam for the full points $\endgroup$ – tien lee Apr 29 '18 at 16:08
  • 1
    $\begingroup$ your professor probably said that whenever a $(-1)^n$ appears, the alternating series test is much likely to work. This does not need that test, the observation that the sequence of terms does not tend to 0 is a simpler argument and enough to prove the divergence. Do not follow instructions as a recipe, understand what you do each time you do it. $\endgroup$ – JustDroppedIn Apr 29 '18 at 16:10

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