## Using the Graph of 1/x, Prove the Harmonic Series Diverges

Problem: Using the graph of $$\frac{1}{x}$$, prove the Harmonic Series diverges. The Harmonic Series is the infinite sum: $\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \cdots$

Solution:

We can draw the harmonic series on the same diagram as $$\frac{1}{x}$$ as the area of several rectangles:

(*diagram)

We can see that the rectangles can be used as an approximation for the area under the graph, but this is an overestimate. Specifically, this tells us that:

$\sum_{n=1}^{\infty} \frac{1}{n}>\int_1^{\infty} \frac{1}{x} \hspace{1mm} dx$

We can evaluate the integral like so:

$\int_1^{\infty} \frac{1}{n}=\lim_{r\to \infty} \int_1^{r} \frac{1}{x} \hspace{1mm} dx$

$=\lim_{r\to \infty} [\ln x]_{x=1}^{x=r}$

$=\lim_{r\to \infty} [\ln r - \ln 1]$

$=\lim_{r\to \infty} \ln r$

But as $$r \to \infty$$ ,  $$\ln r \to \infty$$ (see Lemma below for a proof of this), so the integral is divergent, but we have shown that the integral is less than the harmonic series, and so the harmonic series must diverge.  $$Q.E.D.$$

## Lemma: As $$r \to \infty$$ ,  $$\ln r \to \infty$$

We can prove this by contradiction:

Assume that as $$r \to \infty$$ ,  $$\ln r \to x$$ for some constant x. Then it must also be true that:

$\ln 2r \to x$

$(\ln r + \ln 2) \to x$

But by our assumption, $$\ln r \to x$$, so $$\ln r + \ln 2$$ should tend to $$x+\ln2$$, which is a contradiction. Therefore as $$r \to \infty$$ ,  $$\ln r \to \infty$$.  $Q.E.D.$