Summation to infinity with a general pattern $$
\sum_{n=1}^{\infty} \frac{(n-1)!}{(x+1)(x+2)..... (x+n)}.
$$
I tried solving it by inserting some values
$\frac{1}{x+1}$ +$\frac{1}{(x+1)(x+2)}$+$\frac{2}{(x+1)(x+2)(x+3)}$
$\frac{1}{x+1}$+$\frac{1}{x+1}$-$\frac{1}{x+2}$+$\frac{1}{x+1}$-$\frac{1}{x+2}$+$\frac{1}{x+3}$-$\frac{1}{x+2}$
This gives me a hint that
$\frac{1}{x+1}$ this comes n times
-$\frac{1}{x+2}$ comes n-1times and so on
But since the terms tend to infinity I dont find it useful so any hint?
Answer is $\frac{1}{x}$

A more general result is true.

$\textbf{Lemma:}$ Let $(a_n)_{n\ge 1}$ be a positive sequence. Then for $x \ne 0$,
$$
\sum_{n=1}^{\infty} \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)} = \frac{1}{x}
$$

Proof. Define $A_0 = \frac{1}{x}$ and $$A_N =  \frac{a_1a_2\cdots a_{N}}{x(x+a_1)(x+a_2)\cdots(x+a_N)}$$
Note that $$A_{n-1} - A_n = \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)}$$
So,
$$
\begin{align*}
\sum_{k=1}^{N} \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)} &= \sum_{k=1}^{N} (A_{n-1} - A_n) \\ &= A_0 - A_N \\
&= \frac{1}{x} - \frac{a_1a_2\cdots a_{N}}{x(x+a_1)(x+a_2)\cdots(x+a_N)}
\end{align*}
$$
Letting $N \to \infty$, we get the desired result.
In your case, $a_n = n$.
