Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On Concrete Mathematics section 5.5, which is teaching the hypergeometric functions, generalized factorials is defined as:

\[ \frac 1 {z!} = \lim_{n \to \infty} \binom{n+z}{n}n^{-z} \]


\[ \binom r k = \left\{ \begin{array}{ll} r^{\underline k} / k! = r(r-1) \cdots (r-k+1) / k! & k > 0 \\ 1 & k = 0 \\ 0 & k < 0 \end{array} \right. \]

follows the ordinary definition.

So $z! = z(z-1)!$ for all complex $z$ (except negative integers), then we can check $0! = 1$ and $n! = n(n-1) \cdots 1$ for $n > 0$.

Then, a binomial coefficient can be written

\[ \binom z w = \lim_{\zeta \to z} \lim_{\omega \to w} \frac{\zeta!}{\omega!(\zeta-\omega)!} \]

Let $t_k = \dbinom r k \dbinom s {n-k}$.

However, the succeeding paragraph says that

\[ t_k = \frac{r!}{(r-k)!k!} \frac{s!}{(s-n+k)!(n-k)!}\] and we are no longer too shy to use generalized factorials in these expressions.

without limits (it is said that we must use appropriate limiting values when these formulas give $\infty / \infty$) and considers the ratio $t_{k+1} / t_k$ for all $t_k \neq 0$ and cancels some factorials using the property $z! = z(z-1)!$

I'm "too shy" and my question remains: why can we do such canceling?

To observe closely, we take a variety of an example from section 5.7:

Considering indefinite summation \[ \sum \binom n {-k} \delta k, \qquad n < 0 \]

Let $t(k) = \dbinom n {-k} = \dfrac{n!}{(-k)!(n+k)!}$, we have \[ \frac{t(k+1)}{t(k)} = \frac{n!}{(-k-1)!(n+k+1)!} \frac{(-k)!(n+k)!}{n!} = -\frac{k}{n+k+1} \]

Let $n = -1$, we have $t(k+1) / t(k) = -1$ for $t(k) \neq 0$. But it's wrong for $k = 0$, where $t(1) = 0$ and $t(0) = 1$.

To see how the error happens, we resume the $\lim$ notation:

\begin{align*} t(k+1) &= \binom n {-k-1} \\ &= \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{(n+z_2)!}{(-k-1+z_1)!(n+k+1-z_1+z_2)!} \\ &= \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{-k+z_1}{n+k+1-z_1+z_2} \frac{(n+z_2)!}{(-k+z_1)!(n+k-z_1+z_2)!} \\ &= \binom n {-k} \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{-k+z_1}{n+k+1-z_1+z_2} \end{align*} So when $n = -1$ and $k = 0$, we have $\lim_{z_2 \to 0} \lim_{z_1 \to 0} (-k+z_1)/(n+k+1-z_1+z_2) = 0$ not $-k/(n+k+1)$.

Another example (also from section 5.5) is:

\[ \lim_{x \to -1} \frac{x!}{(x-1)!} = \lim_{x \to -1} x = -1 \] but \[ \lim_{x \to -1} \frac{x!}{(2x)!} = -2 \] because of $(-z)! \Gamma(z) = \pi / \sin(z\pi)$, so expression $(-2)! / (-1)!$ is illegal.

My question is: in the frame of Concrete Mathematics, how to prevent such errors?

Thanks a lot.

share|cite|improve this question
How is $\binom{n+z}{n}$ defined during the definition of $1/z!$ for complex $z$? – anon May 3 '12 at 9:40
@anon Related to the normal definition for binomial coefficents. – Frank Science May 3 '12 at 13:52
@anon I've added something. Fixed. – Frank Science May 3 '12 at 14:01

If the factorials in the limit expression differ by an integer, then cancel away.

But if the difference only becomes an integer in the limit, then you obviously cannot cancel before taking the limit, so you are not allowed to cancel at all.

The difference between $x$ and $2x$ is $x$ which is not an integer until the limit of $-1$ is reached.

share|cite|improve this answer
How to prevent from $\dbinom r k \neq \dbinom r {r-k}$ where $r$ is a negative integer. – Frank Science May 27 '12 at 3:29
@FrankScience If the binomial coefficient is defined by a limit, you don't want to prevent that. The equality is only wrong if you say that binomial coefficients with negative value below is zero. But in the limit definition this is not true anymore. – Phira May 27 '12 at 9:02
The limit definition is always true. When $k$ is a negative integer, $\displaystyle \binom r k = \lim_{u \to r} \lim_{v \to k} \frac {u!} {v!(u-v)!} = \lim_{u \to r} 0 = 0$. The book says that: I see, the lower index arrives at its limit first. That's why $\binom z w$ is zero when $w$ is a negative integer. The value is infinite when $z$ is a negative integer and $w$ is not an integer. – Frank Science May 27 '12 at 9:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.