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Prove that $\ln(x)$ diverges using the fact that the harmonic series diverges.

How can I compare the $\ln$ with the harmonic series, if the harmonic series appears to be more relevant to the derivative of $\ln$?

Edit: show $\ln(x) \rightarrow\infty$ as $x \rightarrow \infty$.

Michael Hardy
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kiwifruit
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    Can you be more precise? A function can't diverge, but we can say that a limit is divergent. I'm guessing that you probably mean to say that $$\lim_{x \to \infty} \ln x = \infty$$ in which case, write down the fact that $\ln x = \int_1^x dt / t$ and estimate. –  Apr 27 '14 at 02:35
  • But how can I use the harmonic series if the harmonic series is a sum, and the integral doesn't have the sum? – kiwifruit Apr 27 '14 at 02:37
  • Draw the area that $\int_1^N dt/t$ represents for large $N$, and fill in some rectangles. –  Apr 27 '14 at 02:39

2 Answers2

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We know that $\int^a_1\dfrac{\mathrm{d}t}{t}=\ln(a)$, and hence, we can use the integral test: $$\sum\dfrac{1}{n}\text{ diverges}\iff\int^\infty_1\dfrac{\mathrm{d}t}{t}\text{ diverges}\\ \sum\dfrac{1}{n}\text{ does diverge, hence }\int^\infty_1\dfrac{\mathrm{d}t}{t}=\lim_{x\to\infty}\ln(x)\text{ diverges.}$$

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Hint: You can use the integral test.

$\sum \frac{1}{n}$ diverges $\iff$ $\int_1^\infty \frac{1}{x} dx$ diverges.

Well, we know the harmonic series $\sum \frac{1}{n}$ diverges, so the integral must diverge. Evaluate it! :)

Kaj Hansen
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