Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals

Theorem

 * $\dfrac {\pi - 3} 4 = \dfrac 1 {2 \times 3 \times 4} - \dfrac 1 {4 \times 5 \times 6} + \dfrac 1 {6 \times 7 \times 8} \cdots$

Proof
Let $f: \R \to \R$ be the real function defined as:


 * $\forall x \in \R: \map f x = x^1 - x^3 + x^5 - x^7 + x^9 - x^{11} + x^{13} - x^{15} \cdots$

We can rewrite this infinite geometric sequence as follows:

Integrating the infinite geometric sequence $3$ times and using Integral of Power, we get:

Integrating the equivalent analytic function $3$ times, we get:

Lemma
We now have:


 * $\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 4} } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} } = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 \paren x^2} 4$

Next we confirm that the infinite geometric sequence on the will converge at $x = 1$.

We are guaranteed convergence by the Alternating Series Test:


 * $\map f 1 = \dfrac 1 {2 \times 3 \times 4} - \dfrac 1 {4 \times 5 \times 6} + \dfrac 1 {6 \times 7 \times 8} + \cdots + \dfrac {\paren {-1}^n } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }$

Finally, we substitute $x = 1$ to obtain our desired result: