Show that the sequence $\{\int_{1}^{n} \frac{\cos t}{t^2} dt\}$ is Cauchy

Question

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?

Answer 1

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

Answer 2

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$$\left|\int_0^m\frac{\cos t}{t^2}\,\mathrm dt-\int_0^n\frac{\cos t}{t^2}\,\mathrm dt\right|\leqslant\left|\frac1n-\frac1m\right|<\frac1N<\varepsilon.$$