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The way I see it you can compare

$$\sum_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) } < 1/n$$

$1/n$ is a $p$-series in which $p = 1 \leq 1 $

So $1/n$ diverges.

Thus $\sum\limits_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) }$ diverges.

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You make a mistake, you are basically saying that since $\sum_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) } < \infty$ it has to be infinite.... But any number is less than infinity... – N. S. Aug 9 '12 at 14:23
No, that's not how the comparison test works. – user5137 Aug 13 '12 at 3:12
up vote 12 down vote accepted

As $\ln(n) > 1$ for $n > e$, $\frac{1}{n^2 \ln(n)} < \frac{1}{n^2}$. The latter is known to converge.

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That makes sense. Can you tell me what the flaw in my argument was ? It still seems like it should work – mr real lyfe Aug 9 '12 at 7:53
Showing that $\sum x$ is bounded by a divergent $\sum y$ says nothing about the convergence of $\sum x$. All series are bounded by divergent ones. The argument is only useful if $y$ converges or $x > y$ instead. – Karolis Juodelė Aug 9 '12 at 7:56
actual flaw is that you have compared to $1/n$ – dato datuashvili Aug 9 '12 at 7:56
@ordinary the flaw in your argument is that you do the comparison using the wrong inequality. To show something diverges with comparison, you need to show that it is at least as large as some divergent series, not smaller than it. – Tobias Kildetoft Aug 9 '12 at 7:56
@ordinary $1/n$ diverges does not imply the series diverges; if your inequality was > then you can say that – pritam Aug 9 '12 at 7:56

Use the Integral test for convergence: Since $\frac d{dt}\text{Ei}(-\log t)=\frac d{dt}\text{li}(\frac1t)=\frac1{t^2\log(t)}$, we get $$ \begin{eqnarray} \int\limits_2^\infty \frac1{n^2\log(n)}dn&=&\text{Ei}(-\log n)\Biggr|_{2}^\infty&<&\infty\\&=&\underbrace{\text{Ei}(-\log \infty)}_{=0}-\text{Ei}(-\log 2)\\ &=&-\text{li}(\frac12)\\ &=&-\int_{0}^{1/2}\frac{dn}{\ln n}&&\hskip0.7in (*) \\ &=&0.378\dots&<&\infty \end{eqnarray} $$ your sum converges: $(*)$ is finite, since $\underbrace{\text{li}(1)}_{-\infty}<\int_{0}^{1/2}\frac{dn}{\ln n}< \underbrace{\text{li}(0)}_{=0}$ and $0<\frac12<1.$

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