Concrete Mathematics: sum evaluates to indeterminate number

Question

On page 58 in Concrete Mathematics [2nd Ed] (Graham, Knuth, Patashnik), one of the sums is evaluated like so:

$$ \sum_{k \geq 0} x^k = \lim_{n \to \infty} \frac{1 - x^{n+1}}{1 - x} = \begin{cases} 1/(1-x), \quad &\text{if} \; 0 \leq x < 1; \\ \infty, \quad &\text{if} \; x \geq 1. \end{cases} $$

It bothers me that when $x = 1$ there is a division by zero; however, the numerator would also equal 0. From Wikipedia,

Since any number multiplied by zero is zero, the expression $\frac{0}{0}$ is also undefined; when it is the form of a limit, it is an indeterminate form.

Why is it OK to leave the solution like the authors did? Or, is it a typo?

Answer 1

You're right, this is slightly sloppy. The expression in the middle isn't valid for $x=1$. Nevertheless, the result on the right is correct, since for $x=1$ we have

$$ \sum_{k\ge0}x^k=\sum_{k\ge0}1=\lim_{n\to\infty}n=\infty\;. $$

Answer 2

First, keep in mind that, when $x$ is close to $1$, $x^n$ is very small (for example, if $x=0.999$, $x^{10000}\approx 4.52\times 10^{-5}$ while the denominator is still finite.

Now, what you could do is to let $x=1-\epsilon$ making $$S_n=\frac{1 - x^{n+1}}{1 - x}=\frac{1-(1-\epsilon)^{n+1}} \epsilon$$ Now, use the binomial expansion or Taylor series to get $$S_n=(n+1)-\frac{n (n+1)}{2}\epsilon +\frac{n (n+1)(n-1)}{6} \epsilon ^2+O\left(\epsilon ^3\right)$$ and see what happens when $\epsilon \to0$ : $S_n\sim (n+1)$