Solution: Let $n > m$. Consider \begin{align*} \left\vert x_n - x_m \right\vert & = \left\vert \int\limits_{1}^n \frac{\cos t}{t^2} \mathrm{d}t - \int\limits_{1}^m \frac{\cos t}{t^2}\mathrm{d}t \right\vert \\ & = \left\vert \int\limits_{1}^m \frac{\cos t}{t^2}\mathrm{d}t + \int\limits_{m}^n \frac{\cos t}{t^2}\mathrm{d}t - \int\limits_{1}^m \frac{\cos t}{t^2}\mathrm{d}t \right\vert \\ & = \left\vert \int\limits_{m}^n \frac{\cos t}{t^2} \mathrm{d}t \right\vert \\ & \leq \int\limits_m^n \frac{1}{t^2}\mathrm{d}t = \frac{1}{m} - \frac{1}{n}. \end{align*}

Since, the sequence $\left( \frac{1}{n} \right) $ is Cauchy, we can find $n_n \in \mathbb{N} $ such that for $n>m\geq n_0$ \[ \frac{1}{m} - \frac{1}{n} \lt \varepsilon . \] For the above $n_0$, and $n \gt m \geq n_0$, we have \[ \left\vert x_n - x_m \right\vert = \frac{1}{m} - \frac{1}{n} \lt \varepsilon . \] Thus, the sequence is Cauchy.