Prove that $\{ x_n \}_{n \in \mathbb{N}}$ is a Cauchy sequence

Question

In $\mathbb{R}$ with the usual metric, define the sequence $\{ x_n \}_{n \in \mathbb{N}}$ for

$$x_n = \displaystyle\int_{1}^{n} \displaystyle\frac{\cos t}{t^2} dt.$$

Prove that $\{ x_n \}_{n \in \mathbb{N}}$ is a Cauchy sequence

My attempt:

We know that every convergent sequence (with limit s, say) is a Cauchy sequence

But. We have that $\displaystyle\int \dfrac{\cos t}{t^2}dt = - \dfrac{\cos t}{t} - \displaystyle\int \dfrac{\sin t}{t} dt$ but the last integral is not solved with elementary functions.

Thanks

Answer 1

Don't try to evaluate the integral; it's not expressible in terms of elementary functions. Instead, consider $|x_n-x_m|$ (and without loss of generality suppose that $m>n$); we have: $$\left|x_n-x_m\right|=\left|\int_0^n\frac{\cos(t)}{t^2}\,dt-\int_0^m\frac{\cos(t)}{t^2}\,dt\right|=\left|\int_n^m\frac{\cos(t)}{t^2}\,dt\right|.$$ Now use the standard estimate $|\int|\leq\int|\cdots|$ to get: $$\left|\int_n^m\frac{\cos(t)}{t^2}\,dt\right|\leq\int_n^m\frac{|\cos(t)|}{t^2}\,dt\leq\int_n^m\frac{dt}{t^2}.$$ Can you control this integral to show that the sequence is Cauchy?

Answer 2

Use that, for $n<m$, $$\left|x_n-x_m\right|=\left|\int_n^m \frac{\cos t}{t^2}dt\right|\leq\int_n^m\frac{|\cos t|}{t^2}dt$$

and then use that $|\cos t|\leq 1.$