Manipulation of factorials

Question

Is it possible to write $$\frac{(n+m)!}{n! m!}$$ as $f(n)g(m)$, where $f(n)$ is some expression in terms of $n$ and $g$ is some expression in terms of $m$?

For some context, I want to determine the domain of convergence of the series $$\sum_{n,m =1}^{\infty} \frac{(n+m)!}{n! m!} x^n y^m.$$

Answer 1

For some integer $k$ you can write

$$m = k-n$$

In order to get

$$\frac{k!}{n!(k-n)!} \equiv \binom{k}{n}$$

In this way sum becomes a double sum

$$\sum_{n = 1}^{+\infty}\sum_{k = n+1}^{+\infty} \binom{k}{n}x^n y^{k-n}$$

Which is straightforward and it does converge to

$$\frac{xy \left(x +y-2 \right)}{(x-1) (y-1) (x+y-1)}$$

Answer 2

Construct the $2\times2$ matrix $M$ whose $(i,j)$ entry is $\binom{i+j}i$: $$ M = \pmatrix{\binom 2 1&\binom 3 1\\\binom 3 2 & \binom 4 2} = \pmatrix{2&3\\3&6}.$$ If the OP's conjecture was true, $M$ would have rank 1.

Answer 3

Let have $|x|<1$ and $|y|<1$ and $|x+y|<1$ just to ensure that this double series is convergent.

$$S=\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} s(n,m)=\sum\limits_{n=0}^{\infty}\sum\limits_{m=0}^{\infty} s(n,m)-\sum\limits_{m=0}^{\infty}s(0,m)-\sum\limits_{n=0}^{\infty} s(n,0)+s(0,0)$$

We can calculate each term separately:

$\displaystyle\sum\limits_{m=0}^{\infty}s(0,m)=\sum\limits_{m=0}^{\infty} y^m=\frac 1{1-y}$

$\displaystyle\sum\limits_{n=0}^{\infty}s(n,0)=\sum\limits_{n=0}^{\infty} x^n=\frac 1{1-x}$

$\displaystyle s(0,0)=1$

For the double sum, we apply the diagonal summation.

$\color{blue}{s(0,0)}+\color{red}{s(0,1)}+\color{green}{s(0,2)}+\cdots\\\color{red}{s(1,0)}+\color{green}{s(1,1)}+s(1,2)+\cdots\\\color{green}{s(2,0)}+s(2,1)+s(2,2)+\cdots\\ \cdots$

$\displaystyle \sum\limits_{n=0}^{\infty}\sum\limits_{m=0}^{\infty} s(n,m)=\sum\limits_{k=0}^{\infty}\sum\limits_{n+m=k} s(n,m)=\sum\limits_{k=0}^{\infty}\sum\limits_{n=0}^{k} s(n,k-n)=\sum\limits_{k=0}^{\infty}\sum\limits_{n=0}^{k} {k\choose n} x^ny^{k-n}=\sum\limits_{k=0}^{\infty} (x+y)^k=\dfrac 1{1-(x+y)}$

Finally regrouping all terms we get :

$S=\dfrac 1{(1-x-y)}-\dfrac 1{(1-x)}-\dfrac 1{(1-y)}+1=\dfrac{xy\,(2-x-y)}{(1-x)(1-y)(1-x-y)}$