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: \[\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \cdots\]



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


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 \(r \to \infty\) ,  \(\ln r \to \infty\).  \[Q.E.D.\]

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