# 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.

• 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... Aug 9, 2012 at 14:23
• No, that's not how the comparison test works.
– user5137
Aug 13, 2012 at 3:12

## 2 Answers

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

• That makes sense. Can you tell me what the flaw in my argument was ? It still seems like it should work Aug 9, 2012 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. Aug 9, 2012 at 7:56
• actual flaw is that you have compared to $1/n$ Aug 9, 2012 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. Aug 9, 2012 at 7:56
• @ordinary $1/n$ diverges does not imply the series diverges; if your inequality was > then you can say that Aug 9, 2012 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 $0<\frac12<1.$