Simplifying $\sum_{y\in\mathbb{N^*}}\frac{x!y!}{(x+y+1)!}$

Question

I'm looking for a way to simplify the following sum: $$\sum_{y\in\mathbb{N^*}}\frac{x!y!}{(x+y+1)!}$$

I've looked up online for some properties that may help like $$(x+y+1)!=(x+1)!(x+y)!$$ but I'm not sure if it's true or not.

Answer 1

$$\sum\limits_{y \in \mathbb{N}^+} \frac{x! y!}{(x+y+1)!} = x! \sum\limits_{y \in \mathbb{N}^+} \frac{y!}{(x+y+1)!}$$

$$= x! \frac{1}{x \Gamma (x+2)} = \frac{(x-1)!}{\Gamma (x+2)} $$

$$ = \frac{(x-1)!}{(x+1)!} = \frac{1}{x(x+1)}$$

Answer 2

The sum $$\sum_{y=0}^n\frac{x!y!}{(x+y+1)!}=\frac{1}{x}-\frac{(n+1)!(x-1)!}{(x+n+1)!}$$ can be proved with a straightforward induction.

Taking limits we have $$\sum_{y=0}^\infty\frac{x!y!}{(x+y+1)!}=\frac{1}{x}-(x-1)!\lim_{n\to\infty}\frac{(n+1)!}{(x+n+1)!}=\frac{1}{x}$$ for $x\ge 1$

Now just substract the first term, since your sum starts from $y=1$.

Answer 3

Here we use a formula for reciprocal binomial coefficients \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}dz\tag{1} \end{align*} which is based upon the beta function. We obtain \begin{align*} \color{blue}{\sum_{y\in\mathbb{N^*}}}\color{blue}{\frac{x!y!}{(x+y+1)!}} &=\sum_{y=0}^{\infty}\frac{1}{x+y+1}\binom{x+y}{y}^{-1}\\ &=\sum_{y=0}^{\infty}\int_{0}^1z^y(1-z)^{x}\,dz\tag{$\to (1)$}\\ &=\int_{0}^1(1-z)^x\sum_{y=0}^{\infty}z^y\,dz\\ &=\int_{0}^1(1-z)^x\frac{1}{1-z}\,dz\\ &=\int_{0}^1(1-z)^{x-1}\,dz\\ &\,\,\color{blue}{=\frac{1}{x}} \end{align*}