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How does one show that$ \displaystyle \prod_{n=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} = \displaystyle \prod_{n=2}^{\infty}\frac{(k-1)(k+1)}{k^{2}+1} =\frac{\pi}{\sinh \pi}$?

The terms do not seem to telescope.

The only approach I can think of is $$ \displaystyle \prod_{n=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} = \displaystyle \lim_{n \to \infty} \prod_{k=2}^{n}\frac{(k-1)(k+1)}{(k-i)(k+i)} = \lim_{n \to \infty} \frac{\Gamma(n) \Gamma(n+2) \Gamma(2-i) \Gamma(2+i)} {2 \Gamma(n+1-i) \Gamma(n+1+i)} $$ $$ = \frac{\pi}{\sinh \pi}\lim_{n \to \infty} \frac{n(n+1)\Gamma^{2}(n)}{\Gamma(n+1-i) \Gamma(n+1+i)} $$

$ \displaystyle \Gamma(n+1-i) \Gamma(n+1+i) = \prod_{j=0}^{n} (j^{2}+1) \frac{\pi}{\sinh \pi}$, but I don't know how that would help me show that the limit evaluates to $1$.

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Apply the complex Stirling formula to show that the limit is 1. –  Gary Aug 19 '13 at 17:10

1 Answer 1

up vote 15 down vote accepted

If you know the product representation

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

it is rather easy.

Setting $z = i$, we obtain

$$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k^2}\right) = 2 \prod_{k=2}^\infty \left(\frac{k^2+1}{k^2}\right).$$

On the other hand,

$$\prod_{k=2}^n \left(\frac{k^2-1}{k^2}\right) = \frac12\cdot \frac32\cdot \frac23\cdot \frac43 \dotsb \frac{n-1}{n}\cdot \frac{n+1}{n} = \frac12\cdot\frac{n+1}{n},$$


$$\prod_{k=2}^\infty \left(\frac{k^2-1}{k^2}\right) = \frac12.$$

Now divide.

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It will be hard to come with a simpler proof (+1) –  O.L. Jul 23 '13 at 8:40
(+1) nice answer. –  Mhenni Benghorbal Jul 23 '13 at 14:15

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