<s>[INST] 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
 [/INST] 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, $$\frac{x^4(1-x)^6}{1-x^4}=-x^6+6 x^5-15 x^4+20 x^3-16 x^2+12x-16+\frac{4 x}{x^2+1}+\frac{16}{x+1}$$
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