Sequence!

Considering $\frac{1}{(n+1)(n+2)(n+3)(n+4)}$ $=\frac{1}{3}(\frac{1}{(n+1)(n+2)(n+3)}-\frac{1}{(n+2)(n+3)(n+4)})$ $=\frac{1}{3}(\frac{1}{f(n)}-\frac{1}{f(n+1)}),\space f(n)=(n+1)(n+2)(n+3)$ The required expression is now equal to $= \sum_{n=0}^{\infty} \frac{1}{3}(\frac{1}{f(n)}-\frac{1}{f(n+1)})$ $= \frac{1}{3} [(\frac{1}{f(0)}- \frac{1}{f(1)}) + (\frac{1}{f(1)}- \frac{1}{f(2)})++++ \infty \space terms$ $=\frac{1}{3}[\frac{1}{f(0)}- \frac{1}{\infty}]$ $=\frac{1}{3}•\frac{1}{f(0)}=\frac{1}{3}•\frac{1}{1•2•3}$ $\Large =\frac{1}{18}$

Hence answer is $\Large \boxed {18}$

One can even generalize this idea to get the sum any number of terms

$Exp. =\sum_0^\infty \dfrac{1}{(n+1)(n+2)(n+3)(n+4)} =\sum_2^\infty \dfrac{1}{(n-1)n(n+1)(n+2)} \\Spliting~into~partial fractions~\\$ $Exp. = \sum_2^\infty \dfrac{1}{6(n-1)} - \sum_2^\infty \dfrac{1}{6(n+2)} - \sum_2^\infty \dfrac{1}{2n} + \sum_2^\infty \dfrac{1}{2(n+1)} ....(1) \\$
$1st~ term~from ~(1)....\sum_2^\infty \dfrac{1}{6(n-1)} = \sum_2^4 \dfrac{1}{6(n-1)} + \sum_5^\infty \dfrac{1}{6(n-1)}. \\ last~ term~from ~above \sum_5^\infty \dfrac{1}{6(n-1)} = \sum_2^\infty \dfrac{1}{6(n+2)} = -2nd~ term~from~(1) \\~and \\$
$3rd ~term~from~(1)~ - \sum_2^\infty \dfrac{1}{2n} =- \sum_2^2 \dfrac{1}{2n} - \sum_3^\infty \dfrac{1}{2n}. \\ last~ term~from ~above \sum_3^\infty \dfrac{1}{2n} = - \sum_2^\infty \dfrac{1}{2(n+1)}=- 4th~term ~from~(1) \\~and \\$
$Exp.~ reduces~ to,\\ Exp. = \sum_2^4 \dfrac{1}{6(n-1)} - \sum_2^2 \dfrac{1}{2n}\\ Exp. = \dfrac{1}{6}*(1 + 1/2 + 1/3) - \dfrac{1}{2}* \dfrac{1}{2} = \dfrac{1}{18}\\ \boxed{18} \\$

Sanjeet Raria's method is out of the box and much better, but I gave the method above as the normal method. I made finding partial fraction a little easy by shifting the summation starting at 0 to one starting at 2. I have not shown how to find partial fraction assuming that, that can be done with out difficulty.

$\frac{1}{(n+1)(n+2)(n+3)(n+4)} = (\frac{1}{n+1}-\frac{1}{n+2})(\frac{1}{n+3} - \frac{1}{n+4})$

$= \frac{1}{6(n+1)} - \frac{1}{2(n+2)} +\frac{1}{2(n+3)} - \frac{1}{6(n+4)}$

So we have $\frac{1}{6} - \frac{1}{2*2} + \frac{1}{2*3} - \frac{1}{6*4}$

$+ \frac{1}{6*2} - \frac{1}{2*3} + \frac{1}{2*4} - \frac{1}{6*5}$

$+ \frac{1}{6*3} - \frac{1}{2*4} + \frac{1}{2*5} -\frac{1}{6*6}+...$

= $\frac{1}{6*3} = \frac{1}{18}$ so $X = \boxed{18}$ .