Necessary condition for convergence of improper integrals

Question

If $\int_A^{+\infty}f(x)\, dx$ is convergent, can we deduce that $\lim_{x\to +\infty} f(x)=0$
If true, is there any limitation to the value of $A$ or $f(x)$? If false, why?
I just started learning calculus, so forgive me for asking such an elementary question. I didn't find relevant information on my textbook.

Answer 1

Let $f:\mathbb R\to\mathbb R$ be given by $$ f(x) = \begin{cases} 2 & \text{if $2<x<2+\frac{1}{8}$} \\ 3 & \text{if $3<x<3+\frac{1}{27}$} \\ 4 & \text{if $4<x<4+\frac{1}{64}$} \\ \vdots \\ n & \text{if $n<x<n+\frac{1}{n^3}$} \\ \vdots \\ 0 & \text{otherwise} \end{cases} $$

Then $\int_0^\infty f(x) \, dx = 2\cdot\frac{1}{8} + 3\cdot\frac{1}{27} + 4\cdot\frac{1}{64} + \cdots = \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots < \infty$ but $\lim_{x\to\infty} |f(x)| = \infty.$

This function is discontinuous, but a continuous function can be created by replacing the "plateaus" with some kind of bumps.