Convergence of $\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}$ WITHOUT using the integral test. Yesterday I was tutoring a high school student on the convergence of infinite series. We encounter the following series on his practice assignment:
$$\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}$$
He's at the precalculus level so he hasn't seen the integral test which would be the standard method for determining that this particular series diverges.
His teacher though gave him a hint that the formula for infinite interest compound was to be used. I assumed that we had to use something along the lines of: 
$$e^x=\lim_{n\to\infty}\left( 1+\frac{x}{n}\right)^n\implies x=\lim_{n\to\infty} n \ln \left(1+\frac{x}{n}\right)$$
I couldn't do it in the end, it takes me to weird places. Is there really a way of determining this series is divergent without using the integral test?
Edit: The tests that the teacher taught my student and thus the only ones allowed are the following:
nth term test, Comparison/limit comparison test, alternating series test, ratio test.
 A: To use the hint you had: notice that
$$\ln\ln(n+1) - \ln\ln n = \ln\frac{\ln(n+1)}{\ln n}
= \ln\left( 1+ \frac{\ln(1+1/n)}{\ln n} \right)$$
Now, the series 
$$
\sum_{n=2}^N (\ln\ln(n+1) - \ln\ln n)
$$
is telescopic and diverges. And by the limit comparison test, the two series
$$
\sum_n \ln\left( 1+ \frac{\ln(1+1/n)}{\ln n}\right), \qquad \sum_n \frac{1}{n\ln n}
$$
have same behavior since $\lim_{n\to\infty}\frac{\ln\left( 1+ \frac{\ln(1+1/n)}{\ln n}\right)}{\frac{1}{n\ln n}} = 1$.
A: Replace $\ln$ by $\log$ base 2, larger than
$$  \frac{1}{2 \cdot 1} + $$
$$  \frac{1}{3 \cdot 2} +   \frac{1}{4 \cdot 2} +  $$
$$  \frac{1}{5 \cdot 3} +   \frac{1}{6 \cdot 3} +   \frac{1}{7 \cdot 3} +   \frac{1}{8 \cdot 3} + $$
$$  \frac{1}{9 \cdot 4} +   \frac{1}{10 \cdot 4} +   \frac{1}{11 \cdot 4} +   \frac{1}{12 \cdot 4} +\frac{1}{13 \cdot 4} +   \frac{1}{14 \cdot 4} +   \frac{1}{15 \cdot 4} +   \frac{1}{16 \cdot 4} + $$
$$  \frac{1}{17 \cdot 5} +   \frac{1}{18 \cdot 5} +   \frac{1}{19 \cdot 5} +   \frac{1}{20 \cdot 5} +\frac{1}{21 \cdot 5} +   \frac{1}{22 \cdot 5} +   \frac{1}{23 \cdot 5} +   \frac{1}{24 \cdot 5} +  \frac{1}{25 \cdot 5} +   \frac{1}{26 \cdot 5} +   \frac{1}{27 \cdot 5} +   \frac{1}{28 \cdot 5} +\frac{1}{29 \cdot 5} +   \frac{1}{30 \cdot 5} +   \frac{1}{31 \cdot 5} +   \frac{1}{32 \cdot 5} +$$

This is bigger than
$$  \frac{1}{2} + \frac{2}{8} +  \frac{4}{24} + \frac{8}{64} + \frac{16}{160} + \dots$$
or
$$   \frac{1}{2} + \frac{1}{4} +  \frac{1}{6} +   \frac{1}{8} + \frac{1}{10} +  \dots$$
A: A variation on @ClementC's insight uses just the inequality $\ln(1+x)\le x$:
$$
\frac1{n\ln n}\ge\frac{\ln\left(1+\frac1n\right)}{\ln n}\ge\ln\left(1+\frac{\ln\left(1+\frac1n\right)}{\ln n}\right).
$$
The RHS leads to a telescoping sum, so the series in question diverges by comparison.
