How can I show that this integral converges?

Question

So here it is: $\int\limits_2^{+\infty}\left(\cos\frac{2}{x}-1\right)dx$. I've tried to use Cauchy, Dirichlet and Abel's tests, but can't seem to figure this out. Mathematica says in converges, but how can I easily show this?

P.S. This is supposed to be easy, but there have been errors in these tasks before, so I don't know if I am being stupid or something is wrong here.

Answer 1

Let $t=1/x$. We then have $dx = - \dfrac{dt}{t^2}$. Hence, $$\int_2^{\infty} \left(\cos(2/x)-1\right)dx = \int_{1/2}^0 \left(\cos(2t)-1\right)\dfrac{-dt}{t^2} = -2\int_0^{1/2}\dfrac{\sin^2(t)}{t^2}dt$$

Now $\sin^2(t)/t^2 \in [0,1]$ for all $t$ and at $t=0$, it can be continuously extended to take on the value of $1$. Hence, the integral clearly exists, since you are integrating a bounded function over a compact interval.

Answer 2

Use the comparison test and compare with $\int_2^\infty\frac{dx}{x^2}$: $$ \lim_{x\to\infty}\frac{\cos\frac2x-1}{1/x^2}=\lim_{t\to0}\frac{\cos(2\,t)-1}{t^2}=-2. $$