Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

My attempt: $$ \begin{align} \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} &= \lim_{n \to \infty} \prod_{k=2}^{n}\frac{(k-1)(k+1)}{(k-i)(k+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n) \Gamma(n+2) \Gamma(2-i) \Gamma(2+i)} {2 \Gamma(n+1-i) \Gamma(n+1+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2) (1-i) \Gamma(1-i) (1+i)i \Gamma(i)}{2 \Gamma(n+1-i)\Gamma(n+1+i)} \\ &= \frac{\pi}{\sinh \pi}\lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} \end{align}$$

I'm not sure how to go about showing that the limit evaluates to $1$.


To evaluate that limit we can use the fact that $ \displaystyle \frac{\Gamma(n)}{\Gamma(n+z)} \sim n^{-z}$ as $ n \to \infty$.

$$ \begin{align} \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} &= \lim_{n \to \infty} \frac{\Gamma(n) (n+1)n \Gamma(n)}{(n-i)\Gamma(n-i) (n+i)\Gamma(n+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)n^{-i}}{\Gamma(n-i)} \frac{\Gamma(n) n^{i}}{\Gamma(n+i)} \frac{n^2+n}{n^{2}+1} \\ &= (1)(1)(1) \\ &= 1 \end{align}$$

share|cite|improve this question
Apply the complex Stirling formula to show that the limit is 1. – Gary Aug 19 '13 at 17:10
up vote 17 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.

share|cite|improve this answer
It will be hard to come with a simpler proof (+1) – Start wearing purple Jul 23 '13 at 8:40
(+1) nice answer. – Mhenni Benghorbal Jul 23 '13 at 14:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.