Convergence of the integral of $\frac{\cos(x)}{x^2}$

Question

For $n$ in the natural numbers let

$$a_n = \int_{1}^{n} \frac{\cos(x)}{x^2} dx$$

Prove, for $m ≥ n ≥ 1$ that $|a_m - a_n| ≤ \frac{1}{n}$ and deduce $a_n$ converges.

I am totally stuck on how to even go about approaching this. All help would be very gratefully received!

Answer 1

$|a_m-a_n| = \left|\displaystyle \int_{n}^m \dfrac{\cos x}{x^2}dx\right| \leq \displaystyle \int_{n}^m \dfrac{|\cos x|}{x^2}dx \leq \displaystyle \int_{n}^m \dfrac{1}{x^2}dx = \dfrac{1}{n} - \dfrac{1}{m} < \dfrac{1}{n}$. Thus $\{a_n\}$ is a Cauchy sequence, hence converges.

Answer 2

We only need to show $\{b_j\}$ is Cauchy, where $b_j=\int_{1}^{j} \frac{\cos x}{x^2} dx$

Proof: $b_j-b_k=\int_{1}^{j} \frac{\cos x}{x^2} dx-\int_{1}^{k} \frac{\cos x}{x^2} dx=\int_{k}^{j} \frac{\cos x}{x^2} dx$

That is, $$|b_j-b_k|=|\int_{k}^{j} \frac{\cos x}{x^2} dx| \leq \int_{k}^{j} |\frac{\cos x}{x^2}| dx$$

(Recall that $|\int_{a}^{b}f(x) dx| \leq \int_{a}^{b}|f(x)| dx$)

We continue our above analysis, $$\leq \int_{k}^{j}\frac{1}{x^2} dx=-\frac{1}{x}\Biggr|_{k}^{j}= -\frac{1}{j}+\frac{1}{k}<\frac{1}{k}$$

Thus, $$|b_j-b_k|<\frac{1}{k}<\epsilon$$,whenever $k>\frac{1}{\epsilon}$, viz., $k>N,N=[\frac{1}{\epsilon}]+1$

To summarize, $\forall \epsilon>0$, let $N=[\frac{1}{\epsilon}]+1$, then whenever $j>k>N$, we have $|b_j-b_k|<\frac{1}{k}<\epsilon$.