Exercise 8.3.1

Question

Supply the details to show that when $x = \pi/2$ the product formula in (2) is equivalent to
$$\frac{\pi}{2} = \lim_{n \to \infty} \left(\frac{2 \cdot 2}{1 \cdot 3}\right) \left(\frac{4 \cdot 4}{3 \cdot 5}\right) \left(\frac{6 \cdot 6}{5 \cdot 5}\right) \cdots \left(\frac{2n \cdot 2n}{(2n-1)(2n+1)}\right),$$
where the infinite product in (2) is interpreted to be a limit of partial products.

Ulisse Mini · Accepted Answer

Plugging $x = \pi/2$ into (2),
$$
1 = \frac{\pi}{2} \prod^\infty_{i=1} \left(1 - \frac{1}{2i}\right)\left(1 + \frac{1}{2i}\right)$$
$$
\frac{2}{\pi} = \prod^\infty_{i=1} \frac{(2i-1)(2i + 1)}{4i^2}
$$
Taking the reciprocal of both sides leads us with the desired equality.

Exercise 8.3.1

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Exercise 8.3.1

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