Prove/disprove that if $f$ is cont, decreasing on $[1, \infty)$, and $\lim_{x \to \infty} f(x) = 0$, then $\int^{\infty}_{1}f(x)dx$ converges.

Question

I think that since the function gets closer and closer to 0, it closes in on an value and thus it does converge. I also know that for $[1,\infty)$, functions $\frac{1}{x^p}$, converge if $p > 1$. How can I put this proof in math language? I have a hard time doing it.

Answer 1

Well, what about $f(x)=\frac{1}{\sqrt{x}}$? It is continuous, decreasing, and $f(x) \to 0$ as $x \to \infty$. Yet the integral does not converge as $$\lim_{x \to \infty}\int_{1}^{x}f(t)dt=\lim_{x \to \infty}(2 \sqrt{x} -2)$$

Note that the same goes for $\frac{1}{x^{p}}$ when $0<p \le 1$.

Your conjecture is false.