As usually, you may replace integration by Lagrange mean value theorem. We have, denoting $f(x)=\log\log x$, $$ a_{n-1}-a_{n}=(f(n)-f(n-1))-\frac1{n\log n}=f'(\theta_n)-\frac1{n\log n}=\frac1{\theta_n\log \theta_n}-\frac1{n\log n},\\ n-1\leqslant \theta_n\leqslant n. $$ So, $a_{n-1}-a_{n}$ is positive, but the series $\sum (a_{n-1}-a_n)$ is dominated by a telescopic series $\sum (f'(n-1)-f'(n))$, this implies that the series $\sum (a_{n-1}-a_n)$ converges, it is equivalent to the fact that $a_n$ converges.

As usually, you may replace integration by Lagrange mean value theorem. We have, denoting $f(x)=\log\log x$, $$ a_{n-1}-a_{n}=(f(n)-f(n-1))-\frac1{n\log n}=f'(\theta_n)-\frac1{n\log n}=\frac1{\theta_n\log \theta_n}-\frac1{n\log n},\\ n-1\leqslant \theta_n\leqslant n. $$ So, $a_{n-1}-a_{n}$ is positive, but the series $\sum (a_{n-1}-a_n)$ is dominated by a telescopic series $\sum (f'(n-1)-f'(n))$, this implies that the series $\sum (a_{n-1}-a_n)$ converges, it is equivalent to the fact that $a_n$ converges.

If you try to avoid also derivatives, I should ask what at all you know about logarithms. If you know somehow, say, that $\log t<t-1$ for $t>0$, you may rewrite this inequality as $$\frac1x< \frac{\log x-\log y}{x-y}<\frac1y$$ for $x>y>0$ (for $t=x/y$, $t=y/x$), hence $$ \frac{\log\log x-\log\log y}{x-y}=\frac{\log\log x-\log\log y}{\log x-\log y}\cdot \frac{\log x-\log y}{x-y} $$ belongs to the interval $(\frac1{x\log x},\frac1{y\log y})$ for $x>y>1$. This is what we really use applying Lagrange theorem in the proof above.