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. $\square$