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. 

Indeed suppose that the statement $\lim_{x\rightarrow \infty} f(x)x^{\alpha+1}=0$ is false. Then, by definition of the limit, there exists a $\varepsilon>0$ such that for every $X\in[0,\infty[$, there exists a $x\geq X$ such that $f(x)x^{\alpha+1}\geq\varepsilon$.
Hence we can construct a sequence $x_1,x_2,x_3,\dots$ in $[0,\infty[$ satisfying: 


*

*$f(x_i) x_i^{\alpha+1}\geq\varepsilon$ for all $i$, and

*$\frac{x_{i}}{x_{i-1}}\geq 2^i$ for all $i$.


Now comes the main idea: Since $f$ is decreasing, we have $f(x)\geq\frac{\varepsilon}{x_i^{\alpha+1}}$ for all $i$ and $x\le x_i$. 
Hence, for all $i\geq 2$, $$\int_{x_{i-1}}^{x_i} f(x) x^\alpha\,\mathrm dx\geq\varepsilon\int_{x_{i-1}}^{x_i} \frac{x^\alpha}{x_i^{\alpha+1}}\,\mathrm dx=\frac\varepsilon{\alpha+1}-\varepsilon\left(\frac{x_{i-1}}{x_i}\right)^\alpha\geq\frac{\varepsilon}{\alpha+1}-\frac{\varepsilon}{2^i}.$$
It follows that $$\int_0^\infty f(x) x^\alpha\,\mathrm dx\geq \sum_{i=2}^\infty \int_{x_{i-1}}^{x_i} f(x) x^\alpha\,\mathrm dx\geq\sum_{i=2}^\infty\frac{\varepsilon}{\alpha+1}-\frac{\varepsilon}{2^i},$$ but the last sum is clearly divergent. Contradiction. $\square$
