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.

  • $\begingroup$ If $0 \le x < 1$, then $\lim_{n \to \infty}x^n = 0$ and you get $\dfrac{1}{1-x}$. If $x \ge 1$ Then $\sum_{k \ge 0}x^k \ge \sum_{k \ge 0}1 \to \infty$ as $k \to \infty$. $\endgroup$ – steven gregory Aug 26 '18 at 18:07
  • $\begingroup$ No techniques the book calls finite calculus is used here, they first write the partial sums in a closed form and take the limit, which converges only for $|x|<1$ $\endgroup$ – Theo Diamantakis Aug 26 '18 at 18:09
  • $\begingroup$ @TheoDiamantakis can you kindly write the procedure $\endgroup$ – Hussain Yousuf Aug 26 '18 at 18:15
  • $\begingroup$ ...they just outlined it for you. Find partial sums then take a limit. $\endgroup$ – Sean Roberson Aug 26 '18 at 18:20
  • $\begingroup$ @SeanRoberson I know how to do it with normal calculus but the authors developed a technique called Finite Calculus it includes difference operator and definite summation, I was wondering did they used that method because the very next example used that $\endgroup$ – Hussain Yousuf Aug 26 '18 at 18:25

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: $$\sum_{k\ge0} x^k = \lim_{n\rightarrow\infty} \frac{1-x^{n+1}}{1-x} = \begin{cases} \frac{1}{1-x} & \text{if $0 \leq x < 1$}, \\ \infty & \text{if $x \geq 1$}. \end{cases}$$


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