Why does $\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}$ diverge if $n \ln(n)$ is greater than $n$ for $n \geq 2$? Why does $\sum_{n=1}^{\infty} \frac{1}{n \ln(n)}$ diverge if $n \ln(n)$ is greater than $n$ for $n \geq 2$.  Shouldn't $\sum_{n=1}^{\infty} \frac{1}{n \ln(n)}$ be comparable to a convergent p-series?  P-series converge for all $p > 1$ and if we try to imagine $n \ln(n)$ as a power of $n$ wouldn't the power be greater than 1?  What am I missing?
 A: First off, you can't start the series $\ \sum_\limits{n}\frac{1}{n\ln n}\ $ at $\ n=1\ $ because $\ \ln 1=0\ $, so $\ \frac{1}{n\ln n}\ $ is undefined for $\ n=1\ $.
The series $\ \sum_\limits{n=2}^\infty\frac{1}{n\ln n}\ $, however, is not comparable to a $\ p$-series for any $\ p>2\ $ because $\ \lim_\limits{n\rightarrow\infty}\frac{n^p}{n\ln n}=\lim_\limits{n\rightarrow\infty}\frac{n^{p-1}}{\ln n}=\infty\ $ for any such $\ p\ $.  Its divergence can be proved by the integral test. The function $\ f(x)=\frac{1}{x\ln x}\ $ is monotone decreasing over the interval $\ [2,\infty)\ $, and
\begin{align}
\int_2^b\frac{1}{x\ln x}\,dx&=\ln\ln b-\ln\ln2\\
&\rightarrow\infty\ \text{ as }\ b\rightarrow\infty\ .
\end{align}
The integral test therefore tells us that $\ \sum_\limits{n=2}^\infty\frac{1}{n\ln n}\ $ diverges.
A: Notice that $f(x)=x\ln(x)$ is increasing for $x\ge 1$, so $g(x)=1/f(x)$ is decreasing. So, using the step functions known from Lebesgue or Riemann integrals, we get
\begin{aligned}
\sum_{n=2}^\infty g(n)
&=\sum_{n=2}^\infty\int_{n}^{n+1}g(n)\mathrm{d}x
\ge\sum_{n=2}^\infty\int_{n}^{n+1}g(x)\mathrm{d}x
=\int_{2}^\infty g(x)\mathrm{d}x\\
&=[\ln(\ln(x))]_{x=2}^{\infty}=\infty.
\end{aligned}
Taking derivatives of $\ln^n(x)$ in the sense of an $n$-fold application of $\ln$, one can see that $(x\prod_{n=1}^\infty\ln^n(x))^{-1}$ marks the barrier, intuitively speaking. So, say $1/(x\ln(x)^{1+\varepsilon})$ (that's a power) is integrable, with anti-derivative $-1/(\varepsilon\ln(x)^{\varepsilon})$.
