This is because they are two different functions. The integral of one does not necessarily have anything to do with the integral of the other. Aside from the integrands both being [math]\dfrac{1}{x^n}[/math] for different n's there is nothing connecting them. You might as well be asking about [math]\int_1^{ \infty } e^{-x} ~ dx[/math].
[math]\int_{1}^{ \infty } \dfrac{1}{x} ~ dx = \lim_{b \to \infty}\int_1^b \dfrac{1}{x} ~ dx = \left . \lim_{b \to \infty }ln|x| \right | _1 ^b[/math] does not converge because [math]\lim_{b \to \infty } ln|b|[/math] does not exist.
[math]\int_{1}^{ \infty } \dfrac{1}{x^2} ~ dx = \lim_{b \to \infty} \int_1^b\dfrac{1}{x^2} ~ dx = - \left . \lim_{b \to \infty } \dfrac{1}{x} \right | _1^b[/math] converges because [math]\lim_{b \to \infty } \dfrac{1}{b}[/math] exists.
If it helps we know that [math]\dfrac{1}{x} \geq \dfrac{1}{x^2}[/math] on the region of integration. So it is at least possible that even though the integral of 1/x is infinite the integral of 1/x^2 is not.
-Dan