Wallis's Product/Proof 1

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Theorem

\(\displaystyle \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}\) \(=\) \(\displaystyle \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2\)


Proof

Into Euler Formula for Sine Function:

\(\displaystyle \dfrac {\sin x} x\) \(=\) \(\displaystyle \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }\)

we substitute $x = \dfrac \pi 2$.


From Sine of Half-Integer Multiple of Pi:

$\sin \dfrac \pi 2 = 1$

Hence:

\(\displaystyle \frac 2 \pi\) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \frac 1 {4 n^2} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac \pi 2\) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {\frac {4 n^2} {4 n^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \frac {\paren {2 n} \paren {2 n} } {\paren {2 n - 1} \paren {2 n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\)


$\blacksquare$


Source of Name

This entry was named for John Wallis.


Sources