1.2 Multiplying Series

Recall the statement of the Algebra of Series:
Theorem 1.3: (Algebra of Series)

Let \(\sum_{n=1}^{\infty}a_n\) and \(\sum_{n=1}^{\infty}b_n\) be convergent series, and let \(\alpha, \beta \in \mathbb{R}\). Then \[\sum_{n=1}^{\infty}(\alpha a_n + \beta b_n) = \alpha\sum_{n=1}^{\infty}a_n + \beta\sum_{n=1}^{\infty}b_n.\]

Now, compare this to the statement of the Algebra of Limits. Notice that we’ve suspiciously omitted any mention about multiplying infinite series together. This is because while for convergent sequences \((x_n)_n\) and \((y_n)_n\), \[\lim_{n\to\infty}(x_ny_n) = \lim_{n\to\infty}x_n \cdot \lim_{n\to\infty}y_n,\] it does not follow that for convergent series \(\sum_{n=1}^{\infty}a_n\) and \(\sum_{n=1}^{\infty}b_n\), \[\sum_{n=1}^{\infty} a_nb_n = \left(\sum_{n=1}^{\infty}a_n\right)\left(\sum_{n=1}^{\infty}b_n\right).\]

If you’re not convinced, we can try this with \(a_n = b_n = \frac{(-1)^{n}}{\sqrt{n}}\). Using the Leibniz test, we can see that both \(\sum_{n=1}^{\infty}a_n\) and \(\sum_{n=1}^{\infty}b_n\) converge, but \[\sum_{n=1}^{\infty} a_nb_n = \sum_{n=1}^{\infty}\frac{1}{n} = \infty.\] So, the next question we need to ask is whether a formula for multiplying convergent series exists, and if so, which series can we apply it to. This question was answered by Cauchy1, and is summarised in the below theorem:

Theorem 1.4: (Cauchy Multiplication Theorem)

Assume that the real series \(\sum_{n=0}^{\infty}a_n\) and \(\sum_{n=0}^{\infty}b_n\) converge absolutely, and for \(n \in \mathbb{N}\), define \[c_n:=\sum_{j=0}^{n}a_j b_{n - j}.\] Then \(\sum_{n=0}^{\infty} c_n\) converges absolutely, and \[\sum_{n=0}^{\infty} c_n = \left(\sum_{n=0}^{\infty}a_n\right)\left(\sum_{n=0}^{\infty}b_n\right).\]

Note that we index the sums starting at \(n=0\) here. This is purely for convenience, and you could equally formulate this theorem for sums starting at \(n=1\) (or any other value of \(n\) for that matter). Furthermore, we require the condition that the two individual sums are absolutely convergent. As you’ve seen in lectures, the Cauchy Multiplication Theorem will fail if the series involved are only conditionally convergent.

  1. If you’re keeping track, this is the third time Cauchy has appeared in this course.↩︎