There's a Gamma in my Pi!

By definition of the Gamma function, $\sum_{n=1}^{\infty}\frac{\Gamma(n)\;\Gamma(n+1)}{\Gamma(n+2)\;\Gamma(n+3)}=\sum_{n=1}^{\infty}\frac{(n-1)!\;n!}{(n+1)!\;(n+2)!}=\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2(n+2)}.$ Decomposition by partial fractions shows that the series can be further reduced to $\sum_{n=1}^{\infty}\left ( \frac{1/2}{n}-\frac{1/2}{n+2}-\frac{1}{(n+1)^2} \right )=\;\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\sum_{n=1}^{\infty}\frac{1}{n^2}-1 \right ).$ The first series on the RHS telescopes giving us

$\frac{1}{2}\left(\frac{3}{2} \right )-\left(\sum_{n=1}^{\infty}\frac{1}{n^2}-1 \right ) =\;\frac{7}{4}-\sum_{n=1}^{\infty}\frac{1}{n^2}=\boxed{\frac{7}{4}-\frac{\pi^2}{6}\approx 0.105}.$

After simplifying, the nth term turns out to be (1/(n)(n+1)(n+1)(n+2)). Which simplifies to (1/(n)-1/(n+1))(1/(n+1)-1/(n+2)). Which after multiplying and simplifying further, turns out to be 0.5×(1/(n) - 1/(n+2)) - (1/n+1)^2. The first term telescopes to 3/4 whereas the second term becomes (π^2)/6 - 1. Hence, the answer is 3/4 - (π^2)/6 +1 = 0.105. Of course upto 3 decimal places.