Evaluating $\sum_{n=1}^{\infty}\frac{(n-1)!}{(x+1)(x+2)(x+3)\cdots(x+n)}$ for positive real $x$ Let $x$ be a positive real number. Evaluate the infinite sum
$$\sum_{n=1}^{\infty}\frac{(n-1)!}{(x+1)(x+2)(x+3)\cdots(x+n)}$$
My approach was writing the expression as
$$ x!\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)\cdots(n+x)}$$
Then I assumed $x=2$ then I got answer as $1/2$, so the answer was $1/x$ (for general). But can I get any idea for a generalized solution of the question?
 A: Partial fractions approach
You can show, via partial fractions:
$$\begin{align}\frac{(n-1)!}{(x+1)\cdots(x+n)}&=\sum_{k=1}^{n}(-1)^{k-1}\frac{\binom{n-1}{k-1}}{x+k}\\&=\int_0^{1}t^x(1-t)^{n-1}\,dt
\end{align}$$
You could also prove this using integration by parts and induction:
$$\int_0^1 t^x(1-t)^{n-1}\,dt =\frac{n-1}{x+1}\int_0^1 t^{x+1}(1-t)^{n-2}\,dt$$
Summing, we have:
$$\begin{align}\sum_{n=1}^{\infty}\frac{(n-1)!}{(x+1)\cdots(x+n)}&=\int_0^1 t^x\cdot \frac1{1-(1-t)}\,dx\\&=\int_{0}^1t^{x-1}\,dx\\&=\frac{1}{x}\end{align}$$
You can switch the sum and integral by the Lebesgue Monotone Convergence Theorem.
Without the monotone convergence theorem, you can just take partial sums:
$$\begin{align}
\sum_{n=1}^{N} &=\int_0^1 t^x\frac{1-(1-t)^N}{t}\,dt\\
&=\frac1x-\int_0^1 t^{x-1}(1-t)^N\,dt\\
&=\frac1x-\frac{N!}{x(x+1)\cdots (x+N)}\\
&=\frac1x-\frac1{x(1+x/1)(1+x/2)\cdots (1+x/N)}
\end{align}$$
And $\prod_{n=1}^N (1+x/n)\geq 1+\sum_{n=1}^N x/n$ diverges to infinity when $x>0,$ so the limit is $\frac1x.$
