Did the authors use finite calculus to evaluate this sum?
If so then how?
$$
\sum_{k\ge0} x^k
= \lim_{n\rightarrow\infty} \frac{1-x^{n+1}}{1-x}
= \begin{cases}
\frac{1}{1-x} & \text{if $0 \leq x < 1$}, \\
\infty & \text{if $x \geq 1$}.
\end{cases}
$$
The book is Concrete Mathematics by Knuth.
Thank you.
If you divide $1-x^{n+1}$ by $1-x$ you will get $\sum_{k=0}^n x^k$
Now if $|x|<1$, then as $n\to \infty $, $x^{n+1}\to 0$
If $x\ge 1$ then the series diverges to $\infty$
Thus the result: $$\sum_{k\ge0} x^k = \lim_{n\rightarrow\infty} \frac{1-x^{n+1}}{1-x} = \begin{cases} \frac{1}{1-x} & \text{if $0 \leq x < 1$}, \\ \infty & \text{if $x \geq 1$}. \end{cases}$$