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.
 A: 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 $f(1)$.
A: I assume both your summation starts from $0$ instead of $1$.
We have
$$a_n = \dfrac1{(2n+1)(2n+2)(2n+3)} = \dfrac1{2(2n+1)} + \dfrac1{2(2n+3)} - \dfrac1{2n+2}$$
This gives us
$$a_n = \dfrac12 \int_0^1x^{2n} dx + \dfrac12 \int_0^1x^{2n+2} dx - \int_0^1 x^{2n+1} dx = \dfrac12 \int_0^1 x^{2n}(1-x)^2dx$$
Hence,
$$\sum_{n=0}^{\infty} a_n = \dfrac12 \int_0^1 \dfrac{(1-x)^2}{1-x^2}dx = \dfrac12 \int_0^1 \dfrac{1-x}{1+x}dx = \int_0^1 \dfrac{dx}{1+x} - \dfrac12 \int_0^1dx = \log(2) - \dfrac12$$
The same idea works for the other series as well and I will leave it to you to work out the details. Be careful on two counts, while following the above technique:
$1$. Note that I wrote $\dfrac1{2n+2}$ as $\displaystyle \int_0^1 x^{2n+1} dx$ and not as $\displaystyle \dfrac12 \int_0^1 x^n dx$. Though both are valid ways to obtain $\dfrac1{2n+2}$, if you do the second way, when you sum it up, you are changing the order of summation and hence will get a different incorrect answer.
$2$. Also, make sure to justify the change of integration and limits.
A: This is similar, but without partial fractions!
$$(1)$$
$$  \begin{aligned}  \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)(2n+3)}  & =\sum_{n\geq 0} \frac{\Gamma(2n+1)}{\Gamma(2n+4)} \\& =\sum_{n\geq 0}\frac{1}{2}\mathrm{B}(2n+1, \,3) \\& = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}x^{2n}(1-x)^2\; dx\\& = \frac{1}{2}\int_{0}^{1} \frac{1-x}{1+x} \;{dx}  \\& = \ln 2-\frac{1}{2} \end{aligned}$$
$$(2)$$
$$  \begin{aligned}  \sum_{n\geq 0}\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}  & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx  \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$
A: \begin{align}
&\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]&=
{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]&=
{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]&=
{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]&=
{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]&
\end{align}
\begin{align}
\sum_{n = 0}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}
&=
\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)}}
&=
\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]&
\end{align}
$\Psi\left(z\right)$ is the Digamma function and
$\gamma = 0.57721566490153286060651209008240243104215933593992\ldots$ is the Euler-Mascheroni constant.
