Proving convergence of sequence $a_n = \int^n_1 \frac{\cos x}{x^2} dx$

Question

I have a sequence $(a_n)$ defined by:

$$a_n = \int^n_1 \frac{\cos x}{x^2} dx$$

I want to prove that this sequence converges, and I have been given a hint:

Prove, for $m \geq n \geq 1$ that $|a_m - a_n| \leq n^{-1}$.

After that is proven, then the sequence is Cauchy and must converge.

My problem is that I cannot prove this. I've tried evaluating this integral:

$$ |a_m - a_n| = \left|\int^m_n \frac{\cos x}{x^2} dx\right|$$

And my hope was that eventually, after application of some known inequalities e.g. $\frac{1}{n} \cos n \leq \frac{1}{n}$ and the triangle law, eventually I'd end up with the result, but I haven't been successful in getting all the terms with $m$ in them to cancel.

Any help would be appreciated.

Answer 1

$\left|\int^m_n \frac{\cos x}{x^2} dx\right|\le$ $\left|\int^m_n |\frac{\cos x}{x^2}| dx\right|\lt \left|\int^m_n \frac{1}{x^2} dx\right|=| \int^m_n d(-\frac{1}{x})|=[-\frac{1}{x}]_n^m=-\frac{1}{m}+\frac{1}{n} $

Hence $(a_n)$ is Cauchy.