Show that the sequence $\{\int_{1}^{n} \frac{\cos t}{t^2} dt\}$ is Cauchy Here is my attempt but I got stuck.
Proof. Let $\varepsilon > 0$. Choose $N$ (any hints for this) and let $m > n \geq N$. Then
\begin{align*}& \left|\int_{1}^{m} \frac{\cos t}{t^2} dt - \int_{1}^{n} \frac{\cos t}{t^2} dt\right|=\\ & = \left|-\int_{m}^{1} \frac{\cos t}{t^2} dt - \int_{1}^{n} \frac{\cos t}{t^2} dt\right| =\\&=\left|-\int_{m}^{n} \frac{\cos t}{t^2} dt\right| = \left|\int_{m}^{n} \frac{\cos t}{t^2} dt\right|=\\ &= \left|\frac{\cos(c)}{c^2}(n-m)\right| = \left|\frac{\cos(c)}{c^2}\right||n-m| =\\&= \frac{|\cos(c)|}{c^2}(m-n)\end{align*}
$\exists c \in (n,m)$ by Mean Value Theorem.
Any hints on how to proceed with this?

Good job with the first step of your attempt. Note that
$$\left|\int_m^n\frac{\cos t}{t^2}dt\right| \le \left|\int_m^n\frac{|\cos t|}{t^2}dt\right| \le \left|\int_m^n\frac{1}{t^2}dt\right| = \left|\frac{1}{n}-\frac{1}{m}\right|$$
