This question is motivated by pedagogical reason, not research. I will provide a simple proof for contrast, but I would like to see another approach that does not involve integrals, instead even more elementary tools.

Prove that the sequence $a_n$ converges if $$a_n=1+\sum_{k=2}^n\frac1{k\log k}-\log\log n.$$ Proof. Rewrite the given sequence as follows $$a_n=1+\sum_{k=2}^n\frac1{k\log k}-\log\log(n+1)+\log\log(n+1)-\log\log n.$$ Since $c_n:=\log\log(n+1)-\log\log n=\log\left(1+\frac{\log(1+1/n)}{\log n}\right)\rightarrow0$, as $n\rightarrow\infty$, we know $a_n$ converges iff $b_n$ converges; where \begin{align*} b_n:&=1+\sum_{k=2}^n\frac1{k\log k}-\log\log(n+1) \\ &=1-\log\log 2+\sum_{k=2}^n\frac1{k\log k}-\int_2^{n+1}\frac{dx}{x\log x}. \end{align*} This allows to compare the integral $\int_2^{n+1}\frac{dx}{x\log x}$ which is dominated by the Upper Riemann sum $\sum_{k=2}^n\frac1{k\log k}$. Therefore, we have positivity of $$d_n:=\sum_{k=2}^n\frac1{k\log k}-\int_2^{n+1}\frac{dx}{x\log x}>0$$ as well as monotonicity $d_{n+1}>d_n$. This becomes more apparent if you draw a picture. Next, notice that $d_n$ represents the excess area between the area under $f(x)=\frac1{x\log x}$ and the Riemann rectangle. Let's estimate $d_n$ by the difference between the Upper and Lower Riemann sums to compute rectangular areas (remember: the line segment $[2,n+1]$ is partitioned throughout by unit segments). Anyways, we obtain $$d_n<\sum_{k=2}^n\left(\frac1{k\log k}-\frac1{(k+1)\log(k+1)}\right)=\frac1{2\log 2}-\frac1{(n+1)\log(n+1)}<\frac1{2\log2}$$ which illustrates boundedness of the sequence $d_n$ and hence that of $b_n$. We know that any increasing sequence bounded from above is convergent. We conclude $b_n$ (and thus $a_n$) is convergent. The proof is complete. $\square$