Why is this a valid proof for the harmonic series?

Question

The other day, I encountered the Harmonic series which I though is an interesting concept. There were many ways to prove that it's divergent and the one I really liked was rather simple but did the job nevertheless. But a proof used integrals and proved that it diverges. But when using integrals, aren't we also calculating fractions with a decimal denominator? I'm not really familiar with calculus and usually only do number theory but doesn't integral find the area under a continuous curve? Thanks in advance for any help! EDIT Here's the integral proof: $$\int_1^{\infty} 1 / x \,dx= {\infty}$$

Answer 1

Consider the following summation:

$$ A=1\cdot 1+\frac12\cdot1+\frac13\cdot1+\frac14\cdot1+... $$

That's a sum of the areas of an infinite number of rectangles of height $\frac1n$, where $n\in\mathbb{N}$, and width 1. Do you agree that's nothing more than the harmonic series? Moreover, the area under the curve $f(x)=\frac{1}{x}$ from $1$ to $\infty$ is clearly less than $A$ and lies totally within $A$ (see the picture below). However, the area under the curve $f(x)=\frac{1}{x}$ from $1$ to $\infty$ doesn't add up to a finite number. It goes to infinity. What should be happening with a piece of area that's even larger than that? Obviously, it must also be infinite. Therefore, the harmonic series diverges.

Answer 2

It works using a simple inequality. If $f(x) \leq M$ for $x \in [a,b]$, then $\int_a^b f(x)\,dx \leq M(b-a)$. This is fairly easy to prove:

$$\int_a^b f(x)\,dx \leq \int_a^b M\,dx = M(b-a).$$

Notice that $1/x < 1/n$ on the interval $[n,n+1]$. Then,

$$\infty = \int_1^\infty \frac{1}{x}dx = \sum_{k=1}^\infty \int_k^{k+1} \frac{1}{x}\,dx \leq \sum_{k=1}^\infty \int_k^{k+1}\frac{1}{k}\,dx = \sum_{k=1}^\infty \frac{1}{k}.$$