Pi as Sum of Sequence of Reciprocal of Product of Three Consecutive Integers

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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 have:

\(\displaystyle \map f x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle x \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x} {x^2 + 1}\) Sum of Infinite Geometric Progression

Integrating 3 times will give us the desired series.

\(\displaystyle \map {F''} x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 4} } {\paren {2 n + 2} \paren {2n + 3} \paren {2n + 4} }\) Repeated use of Integral of Power

Lemma

$\displaystyle \int \int \int \dfrac x {x^2 + 1} \rd x \rd x \rd x = x \, \map \arctan x + \dfrac {\paren {x^2 - 1} \, \map \ln {x^2 + 1} - 3 x^2} 4$

with all integration constants at $0$.


$\Box$

Substitute $x = 1$:

\(\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n } {\paren {2 n + 2} \paren {2n + 3} \paren {2n + 4} }\) \(=\) \(\displaystyle \paren {1} \arctan \paren {1} + \dfrac {\paren {1^2 - 1} \ln \paren {1^2 + 1} - 3 \paren {1}^2} {4}\)
\(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {\paren {-1}^{n + 1} } {\paren {2 n} \paren {2n + 1} \paren {2n + 2} }\) \(=\) \(\displaystyle \arctan \paren {1} - \dfrac {3} {4}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi - 3} {4}\)

$\blacksquare$


Sources