The Divergence of the Harmonic Series
Sequences, Partial Sums, and Counterintuitive Infinity
The harmonic series is one of the most well known series in analysis. It shows that following our intuition alone may lead us to false conclusions. However, once we establish the proof of its divergence, it will be much more apparent where our intuition went wrong.
First a Reminder of Sequences
If you are familiar with sequences, the first sequence you usually study is \(\{x_n\} = \frac{1}{n}\). We know this sequence converges (in limit as \(n \to \infty\)) by an \(\varepsilon\)-\(\delta\) proof.
Definition: A sequence of real numbers converges to a limit \(L\), if:
In our example, our claim would be that if \(\{x_n\} = \frac{1}{n}\), then \(\frac{1}{n} \rightarrow 0\), or equivalently:
The proof would be to find \(n\) as a function of \(\varepsilon\). Starting with \(\left|\frac{1}{n} - 0\right| < \varepsilon\), and since \(n \in \mathbb{N}\), we know \(\frac{1}{n} \ge 0\). We can drop the absolute values:
For any \(\varepsilon > 0\), we can find a corresponding natural number \(n\), such that any point in our sequence past the index \(n\) will be less than a distance of \(\varepsilon\) away from our limit point. As \(\varepsilon\) gets smaller, \(\frac{1}{\varepsilon}\) gets larger, forcing us to go further and further along the sequence to stay within that distance.
When Do Series Converge?
To define the convergence of a series, we must first introduce the notion of a partial sum.
A partial sum \(s_m\) of a series is defined as:
Essentially, it is the sum up to a certain index. A series \(\sum_{n = 1}^{\infty}\{x_n\}\) converges to a value \(L\) if the sequence of its partial sums \(s_m\) converges to \(L\). In simpler terms, for a series to converge, the "additions" must get small enough quickly enough to prevent the sum from growing without bound.
Why is the Harmonic Series Counterintuitive?
Since we know that \(\lim_{n \to \infty} \frac{1}{n} = 0\), it is intuitive to assume that eventually, we are adding up numbers so small they "don't really add anything." However, this sum goes to \(\infty\); it diverges.
Consider the first term, \(n = 1\), giving a value of 1. If we kept adding 1 infinitely, it would clearly diverge. Our goal is to show we can partition the series to keep "producing" 1.
- Terms \(n=2\) to \(4\): \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12} \approx 1\)
- Terms \(n=5\) to \(12\): \(\frac{1}{5} + \dots + \frac{1}{12} \approx 1\)
- Further Terms: We then add the next 22 values (\(n=13\) to \(34\)) to get approximately 1, then the next 60, and so on.
We find that we need at most somewhere between \(2^{k}\) and \(2^{k+1}\) numbers to add up to 1, where \(k\) is the group index. Because we can group the sum in a way that always produces 1, we effectively have an infinite sum of 1's, resulting in divergence to infinity.