If $f:[0,\infty)\to[0,\infty)$ is continuous & decreasing and $α>0$ s.t. $\int_{0}^\infty x^α f(x)dx<\infty$, $\lim_{x\to\infty}f(x)x^{α+1}=0$ Let $f: [0,\infty)\rightarrow [0,\infty)$ be a continuous and decreasing function. Suppose that exists an $\alpha>0$ such that $\int_{0}^{\infty} x^\alpha f(x) dx < \infty$. Prove that $\lim_{x\rightarrow \infty} f(x)x^{\alpha+1}=0$.
First I tried to integrate using parts formula because I thought that was a good way for arriving to the limit. And then I tried using the definition of convergence but I didn't know how to finish. 

We argue by contradiction. If not, then 
$$\lim _{x\to \infty} f(x) x^{\alpha +1} =0$$
is false, and thus there is an $\epsilon_0 >0$ and a sequence $\{s_n \}$ of positive real numbers so that 
$$ f(s_n) s_n^{\alpha +1} \ge \epsilon _0$$
for all $n\in \mathbb N$. 
Now since $\int_0^\infty f(x) x^\alpha dx <\infty$, there is $M>0$ so that 
$$ \int_M ^y f(x) x^\alpha dx < \frac{\epsilon_0}{2(\alpha +1)}$$
for all $y>M$. Since $\{s_n\}$ converges to infinity, for all $x>M$, there is $s_n$ so that $s_n >x$. 
since $f$ is decreasing, 
$$ \int_x^{s_n} f(t) t^\alpha dt \ge  \int_x^{s_n} f(s_n) t^\alpha dt= \frac{f(s_n)}{\alpha +1} ( s_n^{\alpha +1} - x^{\alpha +1})\ge \frac{1}{\alpha +1}( f(s_n)s_n^{\alpha +1} - f(x) x^{\alpha +1}). $$
So 
$$\epsilon_0 \le f(s_n) s_n^{\alpha +1} \le(\alpha +1)  \int_x^{s_n} f(t) t^\alpha dt + f(x) x^{\alpha+1}\le \epsilon_0/2 +  f(x) x^{\alpha+1}$$ 
which implies $f(x) x^{\alpha +1} \ge \epsilon_0/2$ for all $x>M$. 
But then 
$$ \int_M ^y f(x) x^\alpha dx\ge \frac{\epsilon_0}{2} \int_M^y \frac{1}{x} dx = \frac{\epsilon}{2} (\ln y  - \ln M) $$
which is unbounded as $y \to +\infty$. This is a contradiction to the assumption that $\int_0 ^\infty f(x) x^\alpha dx <\infty$. 
