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:
\(\ds \map f x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac x {x^2 + 1}\) | Sum of Infinite Geometric Sequence |
Integrating $3$ times will give us the desired series.
\(\ds \map {F''} x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 4} } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }\) | Repeated use of Integral of Power |
Lemma
- $\displaystyle \iiint \dfrac x {x^2 + 1} \rd x \rd x \rd x = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 x^2} 4$
with all integration constants at $0$.
$\Box$
Substitute $x = 1$:
\(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n} {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }\) | \(=\) | \(\ds \paren 1 \map \arctan 1 + \dfrac {\paren {1^2 - 1} \map \ln {1^2 + 1} - 3 \paren 1^2} 4\) | ||||||||||||
\(\ds \sum_{n \mathop = 1}^\infty \dfrac {\paren {-1}^{n + 1} } {\paren {2 n} \paren {2 n + 1} \paren {2 n + 2} }\) | \(=\) | \(\ds \map \arctan 1 - \dfrac 3 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\pi - 3} 4\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$