Wallis product
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Wallis product can be stated as the infinite product
\[\begin{align} \dfrac\pi 2 &= \prod_{n=1}^\infty \left ( \dfrac{2n}{2n-1} \cdot \dfrac{2n}{2n+1} \right) \\ &= \dfrac21 \times \dfrac23 \times \dfrac43 \times \dfrac 45 \times\dfrac65 \times \dfrac67 \times \dfrac87 \times \dfrac89 \times \cdots . \end{align}\]
Proofs:
The tantalizing statement can be proved by the application of integrals, which is quite unexpected, yet an approach that is quite logical to follow by observing the \(\frac{\pi}2\) on one side.
That in fact is the exact method Wallis used to prove it, by integration, by comparing \( \int_0^\pi \sin^nx \, dx \) for even and odd values of \(n,\) and using the fact that increasing \(n\) by 1 results in a change that decreases the change in \(n\) as \(n\) gets larger and larger.
This proof does not require integration (see this problem Wallis Product Proof).
Given that \( f(x) = \frac {x^2}{x^2 - 1} \), compute
\[ 51\times \big[ f(50)\times f(49)\times \cdots \times f(3)\times f(2) \big].\]