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