Can you generalize this!

I think in the limit asked, f(x) should be in the numerator

@Rudresh Tomar I think your answer is wrong. When I solved (which might be wrong) I got $\infty$

We can write $f(x) = \displaystyle \sum _{n=1}^\infty \dfrac {n! }{(n+x)!} = \dfrac {1}{(1+x)!} \times \begin{pmatrix} 1 + \dfrac {2!}{(x+2)} + \dfrac {3!}{(x+3)(x+2)} + \dfrac {4!}{(x+4)(x+3)(x+2)} + .......\end{pmatrix}$

Now we take your limit $L = \displaystyle \lim_{x\rightarrow \infty} \dfrac {(x+1)!}{f(x)}$

On a bit simplification you limit turns out $L =\displaystyle \lim_{x\rightarrow \infty} \dfrac {((x+1)!)^2}{{\color{#3D99F6}{\begin{pmatrix} 1 + \dfrac {2!}{(x+2)} + \dfrac {3!}{(x+3)(x+2)} + \dfrac {4!}{(x+4)(x+3)(x+2)} + .......\end{pmatrix} = 1}}}$

Clearly the denominator at $x \rightarrow \infty$ turns to be equal to $1.$

Leaving us with $L = \displaystyle \lim_{x \rightarrow \infty} ((x+1)!)^2$ which is definitely $\boxed{\infty}$ (Infinity/Not Defined).

For your answer to be $1$ your limit should be $\boxed{{\color{#20A900}{L = \displaystyle \lim_{x \rightarrow \infty } (1+x)! f(x)}}}$

Kindly change your limit.