An elementary proof for a limit? 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. $\square$
 A: Adapting Fedor's argument, we have 
$$
a_n-a_{n-1}=\frac{1}{n\log(n)}+\log\log(n-1)-\log\log n=
\frac{1}{n\log(n)}+\log\left(\frac{\log(n-1)}{\log n}\right),
$$
and, similarly to the calculation from your post,
$$
\log\left(\frac{\log(n-1)}{\log n}\right)=\log\left(1+\frac{\log\frac{n-1}{n}}{\log n}\right)=\log\left(1+\frac{-\frac{1}{n}+O\left(\frac{1}{n^2}\right)}{\log n}\right).
$$
Finally, 
$$
\log\left(1+\frac{-\frac{1}{n}+O\left(\frac{1}{n^2}\right)}{\log n}\right)=
\frac{-\frac{1}{n}+O\left(\frac{1}{n^2}\right)}{\log n}+O\left(\frac{1}{n^2}\right)=-\frac{1}{n\log(n)}+O\left(\frac{1}{n^2}\right).
$$
It follows that $a_n-a_{n-1}=O\left(\frac{1}{n^2}\right)$, and so $a_n$ converges.
The only thing used here is $\log(1+x)=x+O(x^2)$ for, say, $|x|<1/2$. That can be easily established using any reasonable rigorous definition of logarithm.
A: 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.
