Sum of Sequence of Products of 3 Consecutive Reciprocals

Theorem

 * $\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }$

Proof
We observe that:

Hence: