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If $$\lim_{x\to\infty} f'(x) = \infty$$ prove that $$\int_1^\infty \frac{1}{f(x)}\neq\infty$$ if $f'(x) \geq 1$ and $f(x) \geq1$ for all values of $x$. I'm thinking I can find a way to write it in big O notation to show this, but I have no idea how.

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I don't think the word "ambiguous" means what you think it means. – Gerry Myerson Jan 24 '13 at 1:06
Also, I think it's false --- what happens if $f(x)$ is (some simple variation on $x\log x$? – Gerry Myerson Jan 24 '13 at 1:10
$$\lim_{x\to\infty} f'(x) = \infty$$ is not true for $x\log x$ – user1825464 Jan 24 '13 at 1:20
$(x\log x)' =( 1+\log x)\ \ {\buildrel{x\rightarrow\infty}\over\rightarrow\ \ \infty}$. – David Mitra Jan 24 '13 at 1:26
In addition to David Mitra's observation, the conditions $f'(x), f(x)\geq 1$ are spurious, considering that any nonnegative increasing function can be adjusted via $f(x)+x^2+1$ to satisfy them which will give even more counterexamples. – Alex R. Jan 24 '13 at 1:32

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