Exercise 8.4.23

Question

As a parting shot, use the value for $(1/2)!$ and the Gauss product formula in equation (9) to derive the famous product formula for $\pi$ discovered by John Wallis in the 1650s:
$$\frac{\pi}{2} = \lim_{n \to \infty} \left(\frac{2 \cdot 2}{1 \cdot 3}\right) \left(\frac{4 \cdot 4}{3 \cdot 5}\right) \frac{6 \cdot 6 }{5 \cdot 7} \cdots \left(\frac{2n \cdot 2n}{(2n-1)(2n+1)}\right)$$

Ulisse Mini · Accepted Answer

$$\frac{\sqrt{\pi}}{2} = \lim_{n \to \infty} \sqrt{n} \prod^n_{i=1} \frac{i}{i + 1/2} = \lim_{n \to \infty} \sqrt{n} \prod^n_{i=1} \frac{2i}{2i + 1}$$
$$ \begin{aligned}
    \frac{\pi}{2} &= \lim_{n \to \infty} 2n \left(\prod^n_{i = 1} \frac{2i}{2i + 1}\right)\left(\prod^n_{i = 1} \frac{2i}{2i + 1}\right) = \lim_{n \to \infty} \left(\prod^n_{i = 1} \frac{2i}{2i - 1}\right)\left(\prod^n_{i = 1} \frac{2i}{2i + 1}\right)  \
&= \lim_{n \to \infty} \prod^n_{i=1} \frac{(2i)^2}{(2i-1)(2i+1)}
\end{aligned}
$$

Exercise 8.4.23

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Exercise 8.4.23

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