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. $\square$



An elementary proof for a limit?
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.


