Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.
I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)

$$
\begin{align}
\sum_{k=1}^\infty\frac1{2k(2k+1)(2k+2)}
&=\frac12\sum_{k=1}^\infty\left(\frac1{2k(2k+1)}-\frac1{(2k+1)(2k+2)}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac1{2k}-\frac2{2k+1}+\frac1{2k+2}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac2{2k}-\frac2{2k+1}\right)-\lim_{n\to\infty}\frac12\left(\frac12-\frac1{2n+2}\right)\\
&=1-\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}-\frac14\\[3pt]
&=\frac34-\log(2)
\end{align}
$$
