Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$
An alternative approach to Behrouz' fine one through Euler's beta function. We have: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{n!}{(n+4)!}=\sum_{n\geq 0}\frac{\Gamma(n+1)}{\Gamma(n+5)}&=&\frac{1}{\Gamma(4)}\sum_{n\geq 0}B(4,n+1)\\&=&\frac{1}{6}\int_{0}^{1}\sum_{n\geq 0}x^{n}(1-x)^3\,dx\\&=&\frac{1}{6}\int_{0}^{1}(1-x)^2\,dx\\&=&\frac{1}{6}\int_{0}^{1}x^2\,dx = \frac{1}{6}\cdot\frac{1}{3}=\color{red}{\frac{1}{18}}.\end{eqnarray*}$$
With the same approach it is straightforward to check that: $$ \sum_{n\geq 0}\frac{n!}{(n+k)!} = \color{red}{\frac{1}{(k-1)\cdot(k-1)!}}$$ for any $k\in\mathbb{N}^+$.
The same can be achieved by recognizing in the LHS a multiple of a telescopic series.
Write $$\begin{align} {\frac{n!}{(n+4)!}}&=\frac{1}{(n+1)(n+2)(n+3)(n+4)}=\\&=\frac{A}{(n+1)(n+2)(n+3)}-\frac{B}{(n+2)(n+3)(n+4)}\end{align}$$
which gives $(n+4)A-B(n+1)=1 \implies A=B=1/3$
Then we get a telescoping series:
$$\begin{align}\sum\limits_{n=0}^{\infty }{\frac{n!}{(n+4)!}}=\frac{1}{3}\sum\limits_{n=0}^{\infty }\left({\frac{1}{(n+1)(n+2)(n+3)}}-\frac{1}{(n+2)(n+3)(n+4)}\right)=\frac{1}{18}\end{align}$$
Just a tip; For natural number $n$, $$\sum_{k=1}^n\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{1\cdot 2\cdot \cdots\cdot m} -\frac{1}{(n+1)\cdot (n+2)\cdot \cdots\cdot (n+m)}\right) $$
Now, the given infinite sum is equal to
$$\sum_{k=1}^n \frac{(k-1)!}{(k+3)!}=\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}$$
by plugging in $m=3$ to the tip, we would get
$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3} \left( \frac{1}{1\cdot2\cdot3} -\frac{1}{(n+1)(n+2)(n+3)}\right)$$
Take limit, $n \to \infty$. $$\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3} \left( \frac{1}{1\cdot2\cdot3} -\lim_{n \to \infty}\frac{1}{(n+1)(n+2)(n+3)}\right)$$ $$=\frac{1}{18}$$
How to decompose the fraction; $$\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{k\cdot (k+1)\cdot \cdots\cdot (k+m-1)} -\frac{1}{(k+1)\cdot (k+2)\cdot \cdots\cdot (k+m)}\right)$$