Convergence of $\sum_{n=3}^\infty \frac{1}{n\ln n \ln \ln n}$ I am asked in a exercise to check if the following series is convergent or not:
$$\sum_{n=3}^\infty \frac{1}{n\ln n \ln \ln n}$$
I have no idea how to approach this one. I think that neither the root test nor the ratio test will work so we're left with comparison test. But what can I compare this series to?
 A: We use the integral test. Note that the function \begin{equation*}f(x) = \frac{1}{x \ln x \ln \ln x}\end{equation*} is non-negative for $x \geq 3$ (since $3 \geq 1$ and $\ln 3 \geq 1$). It is clearly integrable, and we claim that it is monotone decreasing for $x \geq 3$. To show this, it is equivalent to show that $g(x) = x \ln x \ln \ln x$ is monotone increasing for $x \geq 3$. Indeed, \begin{align*}g^{\prime}(x) &= \ln x \ln \ln x + x\cdot\frac{1}{x}\ln\ln x + x\ln x\cdot\frac{1}{\ln x}\cdot\frac{1}{x} \\ &= \ln x \ln \ln x + \ln \ln x + 1 \\ & >0\end{align*} since, as already noted, $3 \geq 1$ and $\ln 3 \geq 1$. Having shown that all the conditions for the integral test are applicable, it follows that the series converges if and only if the integral \begin{equation*}\int_{3}^{\infty}\frac{1}{x \ln x \ln \ln x}\mathop{}\!\mathrm{d}x\end{equation*} converges.
Thus, we compute \begin{equation*}\int_{3}^{\infty}\frac{1}{x \ln x \ln \ln x}\mathop{}\!\mathrm{d}x = \left.\left(\ln\ln\ln x\right)\vphantom{\int}\right\rvert_{3}^{\infty}\end{equation*} which doesn't converge: we have $\lim_{x \to \infty}\ln\ln\ln x$ is infinite. Accordingly, the series diverges.
