1.1 Two More Convergence Tests
So far, we’ve seen two tests which allow us (under certain conditions) to determine whether a series is convergent or not. These were the comparison test and d’Alembert’s ratio test. There’s two more tests to cover here. The first is yet another thing named after Cauchy! The second concerns series which have alternating terms. This is due to Leibniz, who — depending on your point of view — is considered to be the inventor of calculus.
1.1.1 Cauchy Condensation Test
This test is very good when the terms of a series involve logarithms, and can also be used to show that \[\sum_{n = 1}^{\infty} \frac{1}{n^{\alpha}} \;\;\text{converges} \Longleftrightarrow \alpha > 1.\]
Assume \((a_n)_n\) satisfies \(a_n \geq 0 \; \forall n \in \mathbb{N}\), and is a decreasing sequence. For \(k \in \mathbb{N}\), define \(b_k := 2^ka_{2^k}\). Then \[\sum_{n = 1}^{\infty} a_n \;\; \text{converges}\; \Longleftrightarrow\; \sum_{k = 1}^{\infty} b_k \;\; \text{converges}.\]
We conclude this (rather short) subsection with a link to an example of the Cauchy condensation test in practice. It’s highly unlikely you’ll ever get something like this in the exam, but the numbers involved are so ridiculous it’s worth including here nonetheless!
1.1.2 Leibniz Alternating Series Test
We’ve stated in tutorials that the series \[\sum_{i=1}^{\infty}\frac{(-1)^n}{n}\] is conditionally convergent — that is, that this sum converges, but \[\sum_{i=1}^{\infty}\left\lvert\frac{(-1)^n}{n}\right\rvert = \sum_{i=1}^{\infty}\frac{1}{n}\] does not. But how do we prove this? The answer to this lies within the following convergence test:
Suppose \((a_n)_{n\in\mathbb{N}}\) is a decreasing sequence tending to \(0\) as \(n \to \infty\). Then \[\sum_{n=1}^{\infty} (-1)^n a_n\] is a convergent series.
Moreover, since the sequence \((a_n)_n\) is decreasing towards \(0\), we can say that the value of this sum lies in between \(-a_1\) and \(a_2 - a_1\).