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.
An alternative approach; compute

$$f(x)=\sum_{n=0}^\infty \frac{x^{2n+3}}{(2n+3)(2n+2)(2n+1)}$$

This has the property that $f'''(x)= \sum_{n=0}^\infty x^{2n} = \frac{1}{1-x^2}$. Integrate once  (partial fractions) to get logarithms. Integrate twice more (by parts) and evaluate at 1 to get the answer $f(1)$.
\begin{align}
&amp;\sum_{n = 0}^{\infty}
{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}
=
{1 \over 4}\sum_{n = 0}^{\infty}
\left({1 \over n + 1/2} - {2 \over n + 1} + {1 \over n + 3/2}\right)
\\[3mm]&amp;=
{1 \over 8}\left\lbrack
\sum_{n = 0}^{\infty}{1 \over \left(n + 1/2\right)\left(n + 1\right)}
-
\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)}
\right\rbrack
\\[3mm]&amp;=
{1 \over 8}\left\lbrack
{\Psi\left(1/2\right) - \Psi\left(1\right) \over 1/2 - 1}
-
{\Psi\left(1\right) - \Psi\left(3/2\right) \over 1 - 3/2}
\right\rbrack
=
{1 \over 4}\left\lbrack - \Psi\left(1 \over 2\right) + 2\Psi\left(1\right)
-
\Psi\left(3 \over 2\right)
\right\rbrack
\\[3mm]&amp;=
{1 \over 4}\left\lbrace
-\Psi\left(1 \over 2\right) + 2\Psi\left(1\right)
-
\left\lbrack 2 + \Psi\left(1 \over 2\right)\right\rbrack
\right\rbrace
=
{1 \over 2}\left\lbrack
-\Psi\left(1 \over 2\right) + \Psi\left(1\right) - 1
\right\rbrack
\\[3mm]&amp;=
{1 \over 2}\left\lbrace
-\left\lbrack-\gamma - 2\ln\left(2\right)\right\rbrack
+
\left(-\gamma\right) - 1
\right\rbrace
=
\ln\left(2\right) - {1 \over 2}
\\[1cm]&amp;
\end{align}

\begin{align}
\sum_{n = 0}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}
&amp;=
\ln\left(2\right) - {1 \over 2}
\\[3mm]
\color{#0000FF}{\large\sum_{n = 1}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}}
&amp;=
\left\lbrack\ln\left(2\right) - {1 \over 2}\right\rbrack - {1 \over 6}
=
\color{#0000ff}{\large \ln\left(2\right) - {2 \over 3}}
\\[1cm]&amp;
\end{align}

$\Psi\left(z\right)$ is the Digamma function and
$\gamma = 0.57721566490153286060651209008240243104215933593992\ldots$ is the Euler-Mascheroni constant.