$\binom{n}{j} \cdot (n - j)$ when $j = n$

Question

I am a bit confused about simplifying $\binom{n}{j} \cdot (n - j)$ when $j = n$.

If we simply plug $j = n$, we get $\binom{n}{n} \cdot 0 = 0$.

But if we first perform a cancellation of the $(n-j)$, w'd get

$$ \frac{n!}{j!(n-j)!} (n-j) = \frac{n!}{j! (n-j-1)!} $$

and then substituting $j = n$, we get

$$ \frac{n!}{n! (n - n - 1)!} = \frac{1}{(-1)!} $$

But this is undefined. I don't see what's wrong with the first result, but I also don't see what's wrong with the second result. Though, we can only take the factorial for numbers $ \geq 0$. So $(n - j)!$ should actually be $(n - j) \cdot (n-j-1) \ldots 1$. So I'm actually not sure what the term $\frac{n-j}{(n-j)!}$ should become for $j = n$. Based on the first expression, I guess it should become zero. But how?

Answer 1

Actually this is defined. First write $$ \frac{1}{(-1)!}=\frac{1}{\Gamma(0)}, $$ where $\Gamma(z)$ is the gamma function. Since $1/\Gamma(x)$ is entire we can find the value with a limit. We have for small $\epsilon$ $$ \frac{1}{\Gamma(\epsilon)}\sim\epsilon+\mathcal O(\epsilon^2). $$ Hence, $$ \frac{1}{(-1)!}=\lim_{\epsilon\to 0}\frac{1}{\Gamma(\epsilon)}=\lim_{\epsilon\to 0}\epsilon+\mathcal O(\epsilon^2)=0. $$

See here for a plot of $1/\Gamma(x)$. You can visually verify this limit makes sense.

Answer 2

The usual proof that $k/k!=1/(k-1)!$ assumes $k!=\prod_{j=1}^kj$ ends with a factor of $k$, i.e. that the product is non-empty, which is equivalent to $k\ge1$. I say "usual proof" because we can impose the regularization $1/(-1)!=0$ with e.g. @AdamHendrickson's technique. Alternatively, a sequence $a_n$ for which $a_n=1/n!$ for $n\ge0$ can be defined for $n<0$ so that $a_{n-1}=na_n$ for all $n\in\Bbb Z$. Then $a_n=0$ for all $n\le-1$.

Answer 3

If one has a rule that $n!\cdot \dfrac 1 {(n-1)!} = n,$ then the case $n=0$ is $$ 0!\cdot\frac 1 {(-1)!} = 0 $$ which implies $\dfrac 1 {(-1)!}=0.$

But this is "formal" rather than logically rigorous.