TAOCP 1.2.5 Exercise 23

Exercise 23. [**] [HM20] Prove (16), given that

$$ \pi z \prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right) = \sin \pi z. $$

x 24. [HM21] Prove the handy inequalities

$$ \frac{n^n}{e^{,n-1}} \le n! \le \frac{n^{n+1}}{e^{,n-1}}, \qquad \text{integer } n \ge 1. $$

[Hint: $1+x \le e^x$ for all real $x$; hence $(k+1)/k \le e^{1/k} \le k/(k-1)$.]

Verified: yes
Solve time: 36s

By Euler’s product formula derived from (13),

$$ z! = \prod_{n=1}^{\infty} \frac{n^{1-z}(n+1)^z}{n+z}, $$

hence

$$ z!(-z)! = \prod_{n=1}^{\infty} \frac{n^{1-z}(n+1)^z}{n+z} \cdot \frac{n^{1+z}(n+1)^{-z}}{n-z} = \prod_{n=1}^{\infty} \frac{n^2}{n^2-z^2}. $$

Therefore

$$ \frac{1}{z!(-z)!} = \prod_{n=1}^{\infty} \left(1-\frac{z^2}{n^2}\right). $$

Using the given identity,

$$ \pi z \prod_{n=1}^{\infty} \left(1-\frac{z^2}{n^2}\right) = \sin \pi z, $$

we obtain

$$ \frac{\pi z}{z!(-z)!} = \sin \pi z. $$

Solving for the product of factorials yields

$$ z!(-z)! = \frac{\pi z}{\sin \pi z}. $$

Thus equation (16) is proved:

$$ \boxed{ z!(-z)! = \frac{\pi z}{\sin \pi z} }. $$

This completes the proof.

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