<s>[INST] Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$ $$\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\frac{5!}{9!}+\frac{6!}{10!}+\cdots$$
This goes up to infinity. Trying finite cases may help.
My Attempt:It seems that it is going to be $\frac{1}{18}$. My calculations show that its going near $\frac{1}{18}$.
 [/INST] 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;
$$\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{k\cdot (k+1)\cdot \cdots\cdot (k+m-1)} -\frac{1}{(k+1)\cdot (k+2)\cdot \cdots\cdot (k+m)}\right)$$
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