# Factorial canceling on expansion of binomial coefficients on Concrete Mathematics

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

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

where

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

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.

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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

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