How do you unfold this summation factor?

Question

This is from Concrete mathematics page 27:

If we apply $s_n = s_{n-1} a_{n-1} / b_n$ recursively, at last we will need to know $s_0$, but how did it disappear in eq. 2.11?

Answer 1

Just apply the relation iteratively: \begin{equation*} s_n = \frac{s_{n-1}a_{n-1}}{b_n} = \frac{\frac{s_{n-2}a_{n-2}}{b_{n-1}}a_{n-1}}{b_n} = \frac{s_{n-2}a_{n-1}a_{n-2}}{b_n}b_{n-1}. \end{equation*} If you continue, eventually you will get to \begin{equation*} s_n = \frac{a_{n-1}a_{n-2}\cdots a_1}{b_nb_{n-1}\cdots b_2}s_0. \end{equation*}

Answer 2

There is no error. You omitted a key part of the sentence:

The relation $s_n=s_{n-1}a_{n-1}/b_n$ can be unfolded to tell us that the fraction $$s_n=\frac{a_{n-1}a_{n-2}\ldots a_1}{b_nb_{n-1}\ldots b_2}\;,\tag{2.11}$$ or any convenient constant multiple of this value, will be a suitable summation factor.

Here $(2.11)$ is actually a definition of $s_n$ for $n\ge 2$: it defines a sequence $\langle s_n:n\ge 2\rangle$ that satisfies the recurrence $s_n=s_{n-1}a_{n-1}/b_n$ and hence can be used as a summation factor, as can any constant multiple of it. In fact, using the convention that an empty product is $1$, it defines $s_n$ for $n\ge 1$, since it makes $s_1=1$.