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 $f(1)$.
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.
This is similar, but without partial fractions!
$$(1)$$
$$ \begin{aligned} \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)(2n+3)} & =\sum_{n\geq 0} \frac{\Gamma(2n+1)}{\Gamma(2n+4)} \\& =\sum_{n\geq 0}\frac{1}{2}\mathrm{B}(2n+1, \,3) \\& = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}x^{2n}(1-x)^2\; dx\\& = \frac{1}{2}\int_{0}^{1} \frac{1-x}{1+x} \;{dx} \\& = \ln 2-\frac{1}{2} \end{aligned}$$
$$(2)$$
$$ \begin{aligned} \sum_{n\geq 0}\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$
$$ \begin{align} &\sum_{n=0}^\infty\frac1{(2n+1)(2n+2)(2n+3)}\\ &=\sum_{n=0}^\infty\left(\frac{1/2}{2n+1}+\frac{-1}{2n+2}+\frac{1/2}{2n+3}\right)\\ &=-\frac{\frac12}{1}+\frac{\frac12}{1}\tag{add $0$}\\ &\hphantom{=-\frac{\frac12}{1}}+\frac{\frac12}{1}-\frac12+\frac{\frac12}{3}\tag{$n=0$}\\ &\hphantom{=-\frac{\frac12}{1}+\frac{\frac12}{1}-\frac12}+\frac{\frac12}{3}-\frac14+\frac{\frac12}{5}\tag{$n=1$}\\ &\hphantom{=-\frac{\frac12}{1}+\frac{\frac12}{1}-\frac12+\frac{\frac12}{3}-\frac14}+\frac{\frac12}{5}-\frac16+\dots\tag{$n=2$}\\ &=\color{#C00000}{-\frac12}+\color{#00A000}{1-\frac12\,+\frac13-\frac14+\frac15-\frac16+\dots}\\ &=\color{#00A000}{\log(2)}\color{#C00000}{-\frac12} \end{align} $$
$$ \begin{align} &\sum_{n=0}^\infty\frac1{(4n+1)(4n+2)(4n+3)(4n+4)}\\ &=\sum_{n=0}^\infty\left(\frac{1/6}{4n+1}+\frac{-1/2}{4n+2}+\frac{1/2}{4n+3}+\frac{-1/6}{4n+4}\right)\\ &=\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)+\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)\\ &\hphantom{\sum_{n=0}^\infty}+\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)\\ &=\color{#00A000}{\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)}+\color{#0000FF}{\sum_{n=0}^\infty\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)}\\ &+\color{#C00000}{\sum_{n=0}^\infty\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)}\\ &=\color{#00A000}{-\frac\pi{24}}\color{#0000FF}{-\frac{\log(2)}{12}}\\ &\color{#C00000}{+\frac{\log(2)}{3}}\\ &=\frac{\log(2)}{4}-\frac\pi{24} \end{align} $$ $\color{#00A000}{\text{Leibniz Series}}$, $\color{#0000FF}{\text{Alternating Harmonic Series}}$, and $\color{#C00000}{\text{Alternating Harmonic Series}}$.