In this article, we will quickly prove that the harmonic series, \(\sum_{n=1}^{\infty} \frac{1}{n}\), diverges.
Does 1/n converge? Proof
We will split the terms of the series as follows:
\[ 1 = 1 \]
\[ \frac{1}{2} = \frac{1}{2} \]
\[ \frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]
\[ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2} \]
And so on. We observe that the sum is greater than:
\[ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots \]
This series diverges to infinity, so the harmonic series must also diverge.
Conclusion
This is a nice example of how we can’t always just use our different tests to figure out if a series converges or diverges. Sometimes we have to do a bit of manipulating, and only then can we use our tests (in this case the Comparison Test). There are so many times when a series looks impossible to solve, but we can turn it into one of the cases that is solvable by our very many tests for convergence or divergence. Good luck on your future math journeys!
