Show $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges. I wish to show $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ diverges. I initially wanted to use the comparison test, but couldn't come up with a series that is obviously less than $\frac{1}{n\ln n}$ that diverges.
So I moved on to the integral test. The problem here is that I need to show that $f(x)=\frac{1}{x\ln x}$ is continuous on $[2,\infty)$, using $\epsilon$-$\delta$ definition of continuity. I've been given theorems that allow me to just assert that $\ln x$ is continuous on the interval, as I know $1$ is continuous, $x$ is continuous, and so if $\ln x$ is continuous then as we have a composition of continuous functions, $\frac{1}{x\ln x}$ will be continuous on $[2,\infty)$. The trouble I'm having is showing $\ln x$ is continuous.
I again got stuck doing this. And now it feels like I've completely over-complicated things. I need to be rigorous when showing that this series diverges, but we've only been given a certain amount of tools to use. We can't use Cauchy's condensation test, and if I wish to use the integral test I have to show that $f(x)$ is monotone (easy) and also that $f(x)$ is continuous (and the only tool we have for that is $\epsilon$-$\delta$).
I've seen that there are very similar questions to this on the site, but they don't particularly help in my case.
Have I missed something? Thanks for your time.
EDIT: Thank you everyone for your help. Much appreciated!

To show the continuity, perhaps use this method to show continuity, rather than the epsilon-delta approach.
For an arbitrary point $y$ in your set, consider any sequence $x_n$ converging to $y$.
But we see that $f(x_n) \to f(y)$, which, by definition, shows that $f$ is continuous, as required.
