Comparing Areas under Curves I remembered back in high school AP Calculus class, we're taught that for a series:
$$\int^\infty_1\frac{1}{x^n}dx:n\in\mathbb{R}_{\geq2}\implies\text{The integral converges.}$$
Now, let's compare $$\int^\infty_1\frac{1}{x^2}dx\text{ and }\int^\infty_1\frac{1}{x^3}dx\text{.}$$
Of course, the second integral converges "faster" since it's cubed, and the area under the curve would be smaller in value than the first integral.
This is what's bothering me:  I found out that 
$$\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx<\int^\infty_{1/2}\frac{1}{x^4}dx$$
Can someone explain to me when is this happening, and how can one prove that the fact this is right?
Thanks!

$$\text{The main reason that: }\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx$$
$$\text{is because although that }\int^\infty_1\frac{1}{x^2}dx>\int^\infty_1\frac{1}{x^3}dx$$
$$\text{remember that }\int^1_{1/2}\frac{1}{x^2}dx<\int^1_{1/2}\frac{1}{x^3}dx\text{.}$$
$$\text{So, in this case: }\int^\infty_1\frac{1}{x^2}dx+\int^1_{1/2}\frac{1}{x^2}dx=\int^\infty_1\frac{1}{x^3}dx+\int^1_{1/2}\frac{1}{x^3}dx,$$
$$\text{which means that: }\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx$$
