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#### ozgunozgur

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- Jun 1, 2020

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- Thread starter ozgunozgur
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- Jun 1, 2020

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- Mar 1, 2012

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$\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$

and $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ is divergent.