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?
Best Answer
Let $x = y/2.$ Then
$$\int_1^\infty\frac{dx}{x} = \int_2^\infty\frac{dy}{y}.$$
That is a contradiction unless both integrals equal $\infty.$