In this post we look at the harmonic series

\displaystyle \begin{array}{rcl} \sum^{\infty}_{n=1}\frac{1}{n}= 1+ \frac{1}{2}+ \frac{1}{3}+\ldots. \end{array}

Students often have difficulty coming to grips with this fact that this series diverges when they encounter it in an elementary calculus course. The result may seem somewhat counter-intuitive at first, since as {n} gets larger, only very small terms get added to the existing sum. The harmonic series serves as an example of a divergent series whose {nth} term tends to zero, thus justifying the statement that the converse of the following theorem does not hold in general.

Theorem If a series {\sum^{\infty}_{n=1} a_n} converges then {\lim_{n\rightarrow \infty}a_n= 0}.

How do we show that the harmonic series diverges? One of the standard proofs presented in many textbooks goes like this.
Let {s_n= \sum^{n}_{k=1}\frac{1}{k}} be the {n}-th partial sum. Observe that

\displaystyle \begin{array}{rcl} s_1 &=& 1+ 0\left(\frac{1}{2}\right)\\ s_2 &=& 1+ \frac{1}{2}= 1+ 1\left(\frac{1}{2}\right)\\ s_4 &=& 1+ \frac{1}{2}+ \left(\frac{1}{3}+ \frac{1}{4}\right)\\ & > & 1+ \frac{1}{2}+ \left(\frac{1}{4}+ \frac{1}{4}\right)= 1+ 2\left(\frac{1}{2}\right)\\ s_8 &=& 1+ \frac{1}{2}+ \left(\frac{1}{3}+ \frac{1}{4}\right)+ \left(\frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}\right)\\ & > & 1+ \frac{1}{2}+ \left(\frac{1}{4}+ \frac{1}{4}\right)+ \left(\frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}\right)= 1+ \frac{3}{2}, \end{array}

and in general,
\displaystyle \begin{array}{rcl} s_{2^{n}} \geq 1+ \frac{n}{2} \end{array}

This implies that {\lim_{n\rightarrow \infty}s_{2^{n}}= \infty} and hence the harmonic series diverges (why?). This proof (which was given by Oresme around 1350) is fairly simple and easy to understand by most students and it makes you wonder if there are any other proofs out there which are more “hands on” and which only use techniques that first-year calculus students can understand. It turns out that there are at least thirty-nine proofs of the divergence of the harmonic series which can be found in the excellent article by Kifowit and Stamps titled The Harmonic Series Diverges Again and Again and by Kifowit titled More Proofs of the Divergence of Harmonic Series.

A proof by Johann Bernoulli

In this section we will discuss a remarkable proof of the divergence of the harmonic series which first appeared in the work of Jakob Bernoulli who generously attributed it to his brother Johann Bernoulli. For a fascinating account of the history behind this proof and more, please look at the following article:

William Dunham, The Bernoullis and the harmonic series, College Mathematics Journal, 18 (1987), 18-23.

Before looking at Bernoulli’s proof, let’s put things in perspective a little bit. The proof was devised around 150 years before the concept of convergence of series was made precise (by Cauchy) via the notion of convergence of partial sums. It is therefore understandable that Bernoulli looked at infinite series in a somewhat “naive” way (at least compared to the modern notion of convergence of a series). Nonetheless, the proof is quite refreshing and ingenious because it is based on, strangely enough, the convergence of the series {\sum^{\infty}_{n=1}\frac{1}{n(n+1)}}. In what follows, we present a modern version of Bernoulli’s proof given in the aforementioned article by Kifowit and Stamps.
The series

\displaystyle \begin{array}{rcl} \frac{1}{2}+ \frac{1}{6}+ \frac{1}{12}+ \frac{1}{20}+\ldots= \sum^{\infty}_{n=1}\frac{1}{n(n+1)}= \sum^{\infty}_{n=1}\left(\frac{1}{n}- \frac{1}{n+1}\right) \end{array}

is a convergent telescoping series whose sum is 1. In a similar manner, it can be shown that {\sum^{\infty}_{n=k}\frac{1}{n(n+1)}= \frac{1}{k}} for {k=1, 2, 3, \ldots}. Bernoulli’s proof is by contradiction. Suppose that the harmonic series converges and suppose that its sum is {S}. Then,
\displaystyle \begin{array}{rcl} S &=& 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \ldots\\ &=& 1+ \frac{1}{2}+ \frac{2}{6}+ \frac{3}{12}+ \frac{4}{20}+ \frac{5}{30}+ \frac{6}{42}+ \frac{7}{56}+ \ldots\\ &=& 1+ \left(\frac{1}{2}+ \frac{1}{6}+ \frac{1}{12}+ \frac{1}{20}+ \ldots\right)+ \left(\frac{1}{6}+ \frac{1}{12}+ \frac{1}{20}+ \frac{1}{30}+ \ldots\right)\\ &+& \left(\frac{1}{12}+ \frac{1}{20}+ \frac{1}{30}+ \frac{1}{42}+\ldots\right)+ \ldots\\ &=& 1+ 1+ \frac{1}{2}+ \frac{1}{3}+ \ldots\\ &=& 1+ S. \end{array}

This final statement {S= 1+ S} is a contradiction. Thus the harmonic series diverges.

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