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Let $f :[a , \infty)\to \mathbb{R}$ a positive and Monotonic function such that $\int_a^\infty f$ converge
prove: $\lim_{x\to\infty}f(x)=0$

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  • $\begingroup$ Prove the contrapositive: Assume $\lim_{x\to \infty}f(x)\neq 0$. Then because $f$ is monotonous and positive, either $\lim_{x \to \infty}f(x) = \infty$ or $\lim_{x\to \infty} f(x)= L > 0$. In each of the cases, show that $\int_a^\infty f(x)dx$ doesn't converge. $\endgroup$ – Arthur Nov 19 '16 at 11:28
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That's an easy calculation… assume $$\lim_{x\to\infty}f(x)\not=0$$ then because f is monotone and positive there exists an $\varepsilon > 0$ s.t. $f>\varepsilon$ so $$\int_a^\infty f \ge \int_a^\infty \varepsilon = \infty$$

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  • $\begingroup$ what if $f$ is decreasing ? $\endgroup$ – ameen ali Nov 19 '16 at 11:46
  • $\begingroup$ That also holds if is decreasing… if f decreases and there is no such an $\varepsilon$ the limit equals 0 due to the fact that then the inf equals the limit $\endgroup$ – Gono Nov 19 '16 at 12:17
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Alternative answer: if $f$ is constant then the result is trivial; wlog $f$ is decreasing, since if it is increasing then the integral can't converge.

By the integral test for convergence, $\sum_{i=a}^{\infty} f(i)$ converges. Therefore $f(i) \to 0$ as $i \to \infty$ over the integers.

But $f$ is monotone, so $f(\lfloor x \rfloor) \geq f(x) \geq f(\lceil x \rceil)$; so we're done by the squeeze theorem.

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