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?)

The $n$-th term is
$$\frac12\left(\frac1{2n}-\frac{2}{2n+1}+\frac1{2n+2}\right).$$
The whole series does not telescope but is
$$\frac12\left(\frac12-\frac23+\frac24-\frac25+\frac26-\frac27+\cdots
\right).$$
This is very similar (not identical) to a well-known series...
