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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?

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  • $\begingroup$ numberphile problems with zero. removable singularity. $\endgroup$ – user645636 Dec 23 '19 at 10:35
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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

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

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  • $\begingroup$ I understand what you're saying. What do you think about @ClaudeLeibovici's answer? $\endgroup$ – Jeremy Lindsay Dec 23 '19 at 10:29
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    $\begingroup$ @JeremyLindsay: That answer shows that $$ \lim_{x\to1}\frac{1-x^{n+1}}{1-x}=n+1\;. $$ That's correct and perhaps something the authors had in mind when they wrote it this way. It might also help you understand the connection between the different cases. But it doesn't change the fact that the expression in the middle is undefined for $x=1$, as you correctly quote from Wikipedia (since it doesn't include the limit $x\to1$). $\endgroup$ – joriki Dec 23 '19 at 10:34
  • $\begingroup$ Ah I see, so the limit as $ x \to 1 $ is the missing piece that was confusing me. Thank you! $\endgroup$ – Jeremy Lindsay Dec 23 '19 at 10:40
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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 $\epsilon \to0$ : $S_n\sim (n+1)$

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  • $\begingroup$ Does this mean indeterminate forms may or may not be defined depending on the situation? For example, are there indeterminate forms that cannot be "fixed" with a trick like $x = 1 - \epsilon$? $\endgroup$ – Jeremy Lindsay Dec 23 '19 at 10:28
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    $\begingroup$ @JeremyLindsay: A quotient where both numerator and denominator are $0$ at $x=x_0$ can have any limit or none for $x\to x_0$. For instance, $\frac{x^2}{\sin x}$ goes to $0$ as $x\to0$, $\frac{\sin x}{x^2}$ goes to $+\infty$ as $x$ approaches $0$ from above but to $-\infty$ as $x$ approaches $0$ from below; $\frac{\sin x\sin \frac1x}x$ has no limit as $x\to0$ and rather oscillates between $-1$ and $1$. $\endgroup$ – joriki Dec 23 '19 at 21:45

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