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Suppose $\int_{1}^\infty f(x)dx$ converges, and$ \frac{f(x)}{x}$ is monotone decreasing on $[1,\infty]$.

How can I prove $\lim_{x \to \infty}xf(x)= 0$ ?

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For $1 < x < x_1$ we have $f(x) > x f(x_1)/x_1$ so (if $x_1 > 2$) $\int_{x_1/2}^{x_1} f(x) \ dx > \int_{x_1/2}^{x_1} x \frac{f(x_1)}{x_1} \ dx = \frac{3}{8} f(x_1) x_1 $ But if $\int_1^\infty f(x)\ dx$ converges, $\int_{x_1/2}^{x_1} f(x)\ dx \to 0$ as $x_1 \to \infty$.

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    +1 Nice. Never occurred to me to use integration limits with constant ratio.2012-12-05