Infinite series summation 
$$\sum^{\infty}_{n=0}\frac{1}{(4n+5)(4n+6)\cdots \cdots (4n+11)}$$

Try: $$\sum^{\infty}_{n=0}\frac{(4n+4)!}{(4n+11)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\frac{(4n+4)!\cdot 6!}{(4n+4+6+1)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\int^{1}_{0}x^{4n+4}(1-x)^6dx$$
$$ \frac{1}{6!}\int^{1}_{0}\left(\sum^{\infty}_{n=0}x^{4n}\right)x^4(1-x)^6dx = \frac{1}{6!}\int^{1}_{0}\frac{x^4(1-x)^6}{1-x^4}dx$$
could some help me to solve it thanks

The telescoping series, as answered by lab bhattacharjee, is the good way to go.
With regard to the integral you properly wrote, using long division and partial fraction decomposition, $$\frac{x^4(1-x)^6}{1-x^4}=-x^6+6 x^5-15 x^4+20 x^3-16 x^2+12x-16+\frac{4 x}{x^2+1}+\frac{16}{x+1}$$
