Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals/Proof 2

Proof
The alternating sum can be written as $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1}} {2 n \paren{2 n + 1} \paren{2 n + 2}}$.

By Partial Fraction Decomposition:


 * $\ds \frac 1 {2 n \paren {2 n + 1} \paren {2 n + 2}} = \frac 1 2 \paren{\frac 1 {2 n} - \frac 2 {2 n + 1} + \frac 1 {2 n + 2}}$

Therefore: