<s>[INST] How to show that $\sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+2)(2n+3)}=\ln(2)-1/2$? How i can prove that 
$$
\sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+2)(2n+3)}=\ln(2)-1/2
$$
And
$$
\sum_{n=1}^{\infty} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{1}{4}\left(\ln(2) - \frac{\pi}{6}\right).
$$
Thanks in advance.
 [/INST] This is similar, but without partial fractions!
$$(1)$$
$$  \begin{aligned}  \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)(2n+3)}  & =\sum_{n\geq 0} \frac{\Gamma(2n+1)}{\Gamma(2n+4)} \\& =\sum_{n\geq 0}\frac{1}{2}\mathrm{B}(2n+1, \,3) \\& = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}x^{2n}(1-x)^2\; dx\\& = \frac{1}{2}\int_{0}^{1} \frac{1-x}{1+x} \;{dx}  \\& = \ln 2-\frac{1}{2} \end{aligned}$$
$$(2)$$
$$  \begin{aligned}  \sum_{n\geq 0}\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}  & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx  \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$
</s>