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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.

  • 0
    No, that's not how [the comparison test](http://en.wikipedia.org/wiki/Comparison_test) works.2012-08-13

2 Answers 2

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

  • 0
    ah I understand now. Thank you all2012-08-09
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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 $0<\frac12<1.$