Question: How to apply Stolz-Cesaro in order to evaluate this sequence as n approaches to $\infty$?

I got to $$\lim\_{k \to \infty} \frac{\ln\left(\frac{k+2}{k+1}\right)}{\ln\left(\frac{k+1}{k}\right)}$$

Answer: 
If $m\geqslant n$, then\begin{align}\left|\int\_0^m\frac{\cos t}{t^2}\,\mathrm dt-\int\_0^n\frac{\cos t}{t^2}\,\mathrm dt\right|&=\left|\int\_n^m\frac{\cos t}{t^2}\,\mathrm dt\right|\\&\leqslant\int\_n^m\left|\frac{\cos t}{t^2}\right|\,\mathrm dt\\&\leqslant\int\_n^m\frac1{t^2}\,\mathrm dt\\&=\frac1n-\frac1m\end{align}and, of course, if $m<n$, then $\left|\int\_0^m\frac{\cos t}{t^2}\,\mathrm dt-\int\_0^n\frac{\cos t}{t^2}\,\mathrm dt\right|\leqslant\frac1m-\frac1n$. So, if $\varepsilon>0$, just take $N\in\Bbb N$ such that $\frac1N<\varepsilon$. Then, if $m,n\in\Bbb N$ and $m,n\geqslant N$, then$$\left|\int\_0^m\frac{\cos t}{t^2}\,\mathrm dt-\int\_0^n\frac{\cos t}{t^2}\,\mathrm dt\right|\leqslant\left|\frac1n-\frac1m\right|<\frac1N<\varepsilon.$$
Good job with the first step of your attempt. Note that
$$\left|\int\_m^n\frac{\cos t}{t^2}dt\right| \le \left|\int\_m^n\frac{|\cos t|}{t^2}dt\right| \le \left|\int\_m^n\frac{1}{t^2}dt\right| = \left|\frac{1}{n}-\frac{1}{m}\right|$$