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I'm a little confused how the area approaches infinite. If $\lim_{x\to\infty}\ln(x+1)-\ln(x)=0$, then the area increases less and less as x approaches ∞. If area under the curve is also finite for small $x$ values and doesn't increase much quickly, shouldn't the area under the curve also be finite?

It's just odd that the area starts out small and increases less and less but still manages to approach $\infty$. Please explain to me how this can all be true or if I'm just wrong somewhere

Simply Beautiful Art
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    It's true that it increases very slowly, and the rate of increase is slower and slower. An equivalent conundrum is that the first derivative of $\ln x$ approaches $0$ as $x \to \infty$, yet $\ln x$ is unbounded. – davidlowryduda Sep 26 '15 at 23:42

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Note that$$\int_{a}^{2a}\frac{1}{x}dx \ge \int_{a}^{2a}\frac{1}{2a}dx = \frac{1}{2} $$

Keith McClary
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Although not quite the same as the integral you have posted, in order to convince myself of the divergence of this integral I consider an approximation of the integral. Consider the harmonic series:

$$\sum_{i=1}^{\infty} \frac{1}{i}$$

This question demonstrates that the harmonic diverges by grouping the terms. The general idea is:

$$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$$ $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{2}$$

and so on.

Adding an infinite number of $\frac{1}{2}$s together diverges, hence the harmonic series, $\sum_{i=1}^{\infty} \frac{1}{i}$, diverges. A similar argument can be made for your integral.

Community
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Dair
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The problem is that $1/x$ does not approach $0$ quite fast enough for the area to be finite. In fact, $$\int_1^\infty \frac 1{x^p}\,dx$$ converges for any real $p>1$.

Tim Raczkowski
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