Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals/Proof 1

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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 can rewrite this infinite geometric sequence as follows:

\(\ds \map f x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n + 1}\)
\(\ds \) \(=\) \(\ds x \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n}\) factoring out an x
\(\ds \) \(=\) \(\ds x \sum_{n \mathop = 0}^\infty \paren {-x^2}^ n\) Power of Power
\(\ds \) \(=\) \(\ds \dfrac x {1 - \paren {-x^2} }\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \dfrac x {1 + x^2 }\)


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

\(\text {(1)}: \quad\) \(\ds \int \map f x\) \(=\) \(\ds \int \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 2} } {2 n + 2}\)
\(\text {(2)}: \quad\) \(\ds \iint \map f x\) \(=\) \(\ds \int \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 2} } {2 n + 2}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 3} } {\paren {2 n + 2} \paren {2 n + 3} }\)
\(\text {(3)}: \quad\) \(\ds \iiint \map f x\) \(=\) \(\ds \int \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 3} } {\paren {2 n + 2} \paren {2 n + 3} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 4} } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }\)


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


Lemma

$\ds \iiint \dfrac x {x^2 + 1} \rd x \rd x \rd x = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 x^2} 4$

with all integration constants at $0$.

$\Box$


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 left hand side 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:

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

$\blacksquare$

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