# 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?

• Think about upper bounds instead of exact evaluation. $|\int f| \leq \int |f|$. And $|\cos(x)| \leq 1$ for all $x$. Consider what the tails of $\int 1/t^2 \, dt$ look like. Apr 18, 2021 at 18:19
• How about $$\left| {\int_m^n {\frac{{\cos t}}{{t^2 }}dt} } \right| \le \int_n^m {\frac{1}{{t^2 }}dt} = \frac{1}{m} - \frac{1}{n}<\frac{1}{m}<\frac{1}{N}.$$
– Gary
Apr 18, 2021 at 18:20
• Damn that's neat! Apr 18, 2021 at 18:22

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 $$m, then $$\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.$$
• After step 4 $\big| \int_{n}^{m}\frac{\cos(t)}{t} dt \big|<\frac{1}{n}-\frac{1}{m}<\frac{1}{n}+\frac{1}{m}<\epsilon$ whenever $n,m>N(\epsilon)>\epsilon/2$ Is it correct? Jul 2, 2021 at 18:41
Good job with the first step of your attempt. Note that $$\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|$$