$\lim\limits_{x\to\infty}xf(x) = 0 $? I've got this task to prove that, given $\forall x \geq a$ :
f is monotonically decreasing,
$ f \geq {0}$
$\int_a^\infty f(x)dx $ converges.
Then $\lim\limits_{x\to\infty}xf(x) = 0 $
I proved successfully that $\lim\limits_{x\to\infty}f(x) = 0$ but I don't know if it helps me anyhow.
 A: We have
$\int_a^{+\infty}f(x)dx $ convergent implies by Cauchy criterion that
$$\lim_{x\to +\infty}\int_x^{2x}f(t)dt=0$$
on the other hand, by the first mean
formula,
$$\forall x\geq a \;\;\exists c_x\in[x,2x]\;:$$
$$ \int_x^{2x}f(t)dt=xf(c_x)$$
$$\implies \forall x\geq a\;\; 0\leq xf(2x)\leq xf(c_x)$$
$$\implies \lim_{x\to+\infty} 2xf(2x)=0$$
qed.
A: Since $f$ is decreasing and non-negative,
$$
\begin{align}
4\int_a^\infty f(x)\,\mathrm{d}x
&=4\sum_{k=0}^\infty\int_{a2^k}^{a2^{k+1}}f(x)\,\mathrm{d}x\\
&\ge4\sum_{k=0}^\infty\overbrace{\ \ \ \ a2^k\ \ \ \ }^\text{width of interval}\ \ \overbrace{f\!\left(a2^{k+1}\right)}^\text{minimum of $f$}\\
&=\sum_{k=0}^\infty a2^{k+2}f\!\left(a2^{k+1}\right)\tag{1}
\end{align}
$$
Therefore, the sum in $(1)$ converges. Thus, the terms must tend to $0$. That is,
$$
\lim_{k\to\infty}a2^{k+2}f\!\left(a2^{k+1}\right)=0\tag{2}
$$
For $x\in\left[a2^{k+1},a2^{k+2}\right]$,
$$
x\,f(x)\le a2^{k+2}f\!\left(a2^{k+1}\right)\tag{3}
$$
Therefore,
$$
\lim_{x\to\infty}x\,f(x)=0\tag{4}
$$
