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.

  • $\begingroup$ Hint: $\left|\int\limits_{n}^{m}{\frac{\cos x}{x^2}\text{ d}x}\right|\le\int\limits_{n}^{m}{\left|\frac{\cos x}{x^2}\right|\text{ d}x}\le\int\limits_{n}^{m}{\frac{1}{x^2}\text{ d}x}$. $\endgroup$ – Joey Zou Nov 28 '15 at 17:17
  • 1
    $\begingroup$ If $m\geq n$ then $\frac{1}{m}\leq\frac{1}{n}$ therefore $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n}$ $\endgroup$ – Liddo Nov 28 '15 at 17:23
  • $\begingroup$ Hmm, that's actually really obvious, I should get more sleep before posting questions, haha. $\endgroup$ – pizzaroll Nov 28 '15 at 17:40

$\left|\int^m_n \frac{\cos x}{x^2} dx\right|\le$ $\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.


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