1.2 Series and Convergence
1.2.1 Definitions
The mathematics in this section is nothing new either — it’s really just sequences in disguise. Given a sequence, one thing we can do is add up all the terms and see what happens. This results in the idea of an (infinite) series:1.2.2 Tests for Convergence
Much like with proving sequence convergence, using the definition each time you want to ‘evaluate’ a series can get tedious really quickly. Therefore, we really want a couple of tests which can prove convergence without too much hassle. The first of these tests involves comparing the sizes of two series, and is aptly known as the comparison test.Let \((a_n)_n\) and \((b_n)_n\) be real sequences, and suppose that there exists a \(M \in \mathbb{N}\) such that \(\lvert a_n \rvert \leq b_n \;\forall n \geq M.\) Then, if \(\sum_{n = 1}^{\infty} b_n\) is convergent, \(\sum_{n = 1}^{\infty} a_n\) is convergent.
Naturally, using this, we can also build a test for divergence to \(\infty\) out of the comparison test too.
Let \((a_n)_n\) and \((b_n)_n\) be real sequences. If there exists a \(M \in \mathbb{N}\) such that \(0 \leq a_n \leq b_n \; \forall n \geq M\), and \(\sum_{n = 1}^{\infty} a_n\) diverges, then \(\sum_{n = 1}^{\infty} b_n\) diverges.
Let \((a_n)_n\) be a real sequence with \(a_n \neq 0 \; \forall n \in \mathbb{N}\). Suppose \[\lim_{n\to\infty}\frac{\lvert a_{n+1}\rvert}{\lvert a_n\rvert} = r.\] Then:
- If \(0 \leq r < 1\), \(\sum_{n = 1}^{\infty} a_n\) converges.
- If \(r > 1\), then \(\sum_{n = 1}^{\infty} a_n\) diverges.
- If \(r = 1\), the test is inconclusive.
Like the Growth Factor Test, d’Alembert’s test fails if the terms of the series are formed of ratios of polynomials.
The final test presented here is applicable when the terms of the series alternate in sign: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. Further, the value of this series lies between \(-a_1\) and \(a_2 - a_1\).