# Why does $\sum\limits_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) }$ converge?

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

As $\ln(n) > 1$ for $n > e$, $\frac{1}{n^2 \ln(n)} < \frac{1}{n^2}$. The latter is known to converge.
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 $1/n$ 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 $0<\frac12<1.$