TITLE: How to show $\ \int_1^\infty\frac1xdx\ $ diverges (not using the harmonic series)?
QUESTION [4 upvotes]: I was reading up on the harmonic series,
$H=\sum\limits_{n=1}^\infty\frac1n$, on Wikipedia, and it's divergent, as can be shown by a comparison test using the fact that
$\begin{aligned}H&=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)+\cdots\\&\geq 1+\frac12+\left(\frac14+\frac14\right)+\left(\frac18+\frac18+\frac18+\frac18\right)+\cdots\\&=1+\frac12+\frac12+\frac12+\cdots,\end{aligned}$
where the expression on the right clearly diverges.
But after this proof idea was given, the proof idea using the integral test was given. I understand why $H_n=\sum_{k=1}^n\frac1k\geq \int_1^n \frac{dx}x$, but how is it shown that $\int_1^\infty \frac{dx}x$ is divergent without using the harmonic series in the following way:
$H_n-1\leq \int_1^n \frac{dx}x\leq H_n$, and then using this in the following way, by comparison test:
$\lim\limits_{n\to\infty}H_n=\infty\implies\lim\limits_{n\to\infty}(H_n-1)=\infty\implies\lim\limits_{n\to\infty}\int_1^n \frac{dx}x=\infty$.
So to summarize, is there a way to prove that  $\int_1^\infty \frac{dx}x$ without using the fact that $H$ diverges?

REPLY [2 votes]: Note that 
$$\begin{align}
\int_1^{2^n}\frac{1}{u}\,du&=\int_1^2\frac{1}{u}\,du+\int_2^4\frac{1}{u}\,du+\int_4^8\frac{1}{u}\,du+\cdots +\int_{2^{n-1}}^{2^n}\frac{1}{u}\,du\\\\
&\ge \left(\frac{1}{1}\right)\left(2-1\right)+\left(\frac{1}{2}\right)\left(4-2\right)+\left(\frac{1}{4}\right)\left(8-4\right)+\cdots +\left(\frac{1}{2^{n-1}}\right)\left(2^n-2^{n-1}\right)\\\\
&=1+1+1+\cdots +1\\\\
&=n
\end{align}$$
Therefore, we find that 
$$\int_1^{2^n}\frac{1}{u}\,du\ge n \to \infty \,\,\text{as}\,\,n\to \infty$$
And we are done!
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