# Chaitanya's Random Pages

## August 5, 2011

### A Collection of Infinite Products – I

Filed under: mathematics — ckrao @ 12:25 pm

This post and the next one show some cool infinite products (mostly taken from [1] and [2]) with an attempt to demystify most of them. A few ideas are mentioned, but they are not necessarily the approaches with which the formulas were discovered in the first place! If you wish to continue to be captivated by the mystery, admire the formulas as they are and no need to proceed to the proof section.🙂

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

(2) $\displaystyle \prod_{n=3}^{\infty} \left(1 - \frac{4}{n^2}\right) = \frac{1}{6}$

(3) In general $\displaystyle \prod_{n=k+1}^{\infty} \left(1 - \frac{k^2}{n^2}\right) = \frac{1}{\binom{2k}{k}}$

(4) $\displaystyle \prod_{n=2}^{\infty} \frac{n^3 -1}{n^3 + 1} = \frac{2}{3}$

(5) $\displaystyle \frac{1}{\Gamma(z)} = \lim_{n \rightarrow \infty} \frac{z(z+1)\ldots (z+n)}{n! n^z}= ze^{\gamma z} \prod_{n=1}^{\infty} \left[\left(1 + \frac{z}{n}\right) e^{-z/n}\right] = z\prod_{n=1}^{\infty} \frac{1 + z/n}{\left(1 + 1/n\right)^z}$

Here $\displaystyle \Gamma (z) = \int_0^{\infty} u^{z-1}e^{-u}\ du$ generalises the factorial function, so that $\Gamma (n+1) = n!$ for n a non-negative integer. Also $\displaystyle \gamma := \lim_{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{1}{k} - \log n\right) \approx 0.577$ is the Euler-Mascheroni constant. The first equality in (5) is due to Gauss, the second to Weierstrass, the third to Euler.

(6) $\displaystyle \prod_{n=0}^{\infty} \frac{(n+a_1)(n+a_2)}{(n+b_1)(n+b_2)} = \frac{\Gamma (b_1) \Gamma (b_2)}{\Gamma (a_1) \Gamma (a_2)}$, where $a_1 + a_2 = b_1 + b_2$.

(7) Wallis (1655): $\displaystyle \prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots = \frac{\pi}{2}$

(8) $\displaystyle \prod_{n=1}^{\infty} \left(1 - \frac{1}{(2n)^2}\right) = \frac{2}{\pi}$

(9) $\displaystyle \prod_{n=1}^{\infty} \left(1 - \frac{1}{(2n+1)^2}\right) = \frac{\pi}{4}$

(10) $\displaystyle \sin \pi z = \pi z \prod_{n=0}^{\infty}\left(1 - \frac{z^2}{n^2}\right)$

(11) $\displaystyle \cos \pi z = \prod_{n=-\infty}^{\infty}\left(1 + \frac{z}{n - 1/2}\right) = \prod_{n=0}^{\infty}\left(1 - \frac{4z^2}{(2n+1)^2}\right)$

(12) $\begin{array}{lcl} \prod_{n=0}^{\infty} \left(1 + \frac{(-1)^n}{2n+1} \right) &=& \left(1 + \frac{1}{1}\right) \left(1 - \frac{1}{3}\right) \left(1 + \frac{1}{5}\right) \left(1 - \frac{1}{7}\right) \ldots\\ &=& \frac{2}{1}\cdot \frac{2}{3}\cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots\\ &=& \sqrt{2} \end{array}$
(From Wallis’s product (7) remove fractions containing multiples of 4 and we go from $\pi/2$ to $\sqrt{2}$!)

(13) $\displaystyle \prod_{n=-\infty}^{\infty} \frac{n-z_1}{n-z_2} = \frac{\sin \pi z_1}{\sin \pi z_2}$

(14) $\displaystyle \prod_{n=1}^{\infty} \left(1 +\frac{1}{n^2}\right) = \frac{1}{\pi} \sinh \pi$

(15) $\displaystyle \prod_{n=1}^{\infty} \left(1 +\frac{1}{n^3}\right) = \frac{1}{\pi} \cosh \left(\frac{\pi \sqrt{3}}{2}\right)$

(16) $\displaystyle \prod_{n=2}^{\infty} \left(1 -\frac{1}{n^3}\right) = \frac{1}{3\pi} \cosh \left(\frac{\pi \sqrt{3}}{2}\right)$

#### Proofs:

A. Telescoping products

The idea is that terms cancel and only early and late terms remain, then take the limit as n goes to infinity. For example, to prove (1) we have

$\begin{array}{lcl} \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right) &=& \lim_{N \rightarrow \infty} \prod_{n=2}^N \frac{n-1}{n} \prod_{n=2}^N \frac{n+1}{n}\\ & = & \lim_{N \rightarrow \infty} \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{N-1}{N} \right) \left(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \ldots \cdot \frac{N+1}{N} \right) \\ & =& \lim_{N \rightarrow \infty} \frac{1}{N} \frac{N+1}{2}\\ &=& \frac{1}{2}. \end{array}$

Equations (2) and (3) are proved similarly. To prove (4) we have

$\begin{array}{lcl} \prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1} &=& \lim_{N \rightarrow \infty} \prod_{n=2}^{N}\frac{(n-1)(n^2+n + 1)}{(n+1)(n^2-n + 1)}\\ &=& \lim_{N \rightarrow \infty} \prod_{n=2}^{\infty} \frac{n-1}{n+1} \prod_{n=2}^{\infty} \frac{n(n+1) + 1}{(n-1)n + 1}\\&=& \lim_{N \rightarrow \infty} \left(\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \ldots \cdot \frac{N+1}{N+1} \right)\left(\frac{2\times 3+1}{1\times 2+1} \cdot \frac{3\times 4+1}{2\times 3+1} \cdot \frac{5\times 4+1}{3\times 4+1} \cdot \ldots \cdot \frac{N(N+1)+1}{(N-1)N+1} \right)\\&=& \lim_{N \rightarrow \infty} \frac{1 \times 2}{N(N+1)} \cdot \frac{N(N+1)+1}{1\times 2+1}\\&=& \frac{2}{3}.\end{array}$

B. Use of the Gamma function:

By induction we may prove $\displaystyle \int_0^1 t^{z-1} (1-t)^n\ dt = \frac{n!}{z(z+1)\ldots (z+n)}$. Then setting t=u/n and letting n tend to infinity we arrive at the first equality in (5):

$\displaystyle \Gamma (z) = \int_0^{\infty} u^{z-1}e^{-u}\ du = \lim_{n \rightarrow \infty} \frac{n^{z-1}n!}{z(z+1)\ldots (z+n-1)}$

To prove the second equality in (5) refer to this post from the Abstract Nonsense blog. The third equality follows similarly from the definition of the exponential and $\gamma$.

From the first equality of (5) we arrive at (6):

$\displaystyle \prod_{n=0}^{\infty} \frac{(n+a_1)(n+a_2)}{(n+b_1)(n+b_2)} = \frac{\Gamma (b_1) \Gamma (b_2)}{\Gamma (a_1) \Gamma (a_2)}$,

where $a_1 + a_2 = b_1 + b_2$.

In general$a_1, a_2, b_1, b_2$ can be any algebraic expressions and (7) easily generalises to a product of more than two terms.

Setting $a_1, a_2, b_1, b_2$ to 1, 3, 2 and 2 respectively, gives equation (1). Setting $a_1, a_2, b_1, b_2$ to 1, 1, 1/2, 3/2 respectively gives Wallis’s formula (7):

$\displaystyle \prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \prod_{n=0}^{\infty} \frac{(n+1)^2}{(n+1/2)(n+3/2)} = \frac{\Gamma (1/2) \Gamma(3/2)}{\Gamma (1) \Gamma (1)} = \frac{\sqrt{\pi}\times \sqrt{\pi}/2}{1\times 1} = \frac{\pi}{2}$

C. Trigonometric/Hyperbolic results

Formula (7) can also be proved via the infinite product expansion for $\sin \pi z$ (10) shown in my previous mathematical blog post on Tannery’s theorem.

$\displaystyle \sin \pi z = \pi z \prod_{n=0}^{\infty}\left(1 - \frac{z^2}{n^2}\right)$

By a similar approach the infinite product expansion for $\cos \pi z$ (11) can be derived. These two formulas and the expansion of the gamma function (5) can also be derived by application of the Weierstrass factorization theorem.

Setting z to 1/2 in (10) leads to (7). Formula (8) is simply the reciprocal of (7), while (9) is obtained via (6).  It is interesting to compare (1), (8) and (9) to see how different their values can be.

Setting z to 1/4 in the sine formula (10) and then taking reciprocals yields

$\begin{array}{lcl} \sqrt{2} &=& \prod_{n=0}^{\infty}\left(\frac{(2n+1)^2}{(2n+1)^2 - 1/4}\right)\\ &=& \prod_{n=0}^{\infty}\left(\frac{2(2n+1).2(2n+1)}{(2(2n+1)-1).(2(2n+1) +1}\right)\\&=& \frac{2}{1}\cdot \frac{2}{3}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{10}{9}\cdot \frac{10}{11}\cdot\ldots \end{array}$

Formula (13) results from direct application of (10) (a ratio of sines). The hyperbolic results (14)-(16) result partly from (10) and (11) (via $\cosh z = \cos iz, \sinh z = -i \sin iz$) and partly from Wallis’s formula (7).  To prove (15) ((16) is proved similarly) we firstly note that

$\displaystyle \cosh \left( \frac{\sqrt{3} \pi}{2} \right) = \prod_{n=1}^{\infty} \left(1 + \frac{3}{(2n-1)^2}\right).$

Then

$\begin{array}{lcl} \prod_{n=1}^{\infty} \left(1 + \frac{1}{n^3} \right) &=& \prod_{n=1}^{\infty} \frac{(n+1)(n^2-n+1)}{n^3}\\&=& \prod_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{4(n^2 - n + 1)}{4n^2}\\&=& \prod_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{(2n-1)^2(2n+1)}{(2n+1)4n^2} \cdot \frac{4(n^2 - n + 1)}{(2n-1)^2}\\&=&\prod_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{2n-1}{2n+1} \cdot \frac{4n^2-1}{4n^2} \cdot \frac{4(n^2 - n + 1)}{(2n-1)^2}\\&=&\lim_{N \rightarrow \infty}\prod_{n=1}^{N} \frac{n+1}{n} \cdot \frac{2n-1}{2n+1} \cdot \frac{2}{\pi} \cdot \prod_{n=1}^{\infty} \left(1 + \frac{3}{(2n-1)^2}\right)\\&=& \lim_{N \rightarrow \infty} \frac{N+1}{2N+1} \cdot \frac{2}{\pi} \cdot \cosh \left( \frac{\sqrt{3} \pi}{2} \right)\\ &= & \frac{1}{\pi}\cosh \left( \frac{\sqrt{3} \pi}{2}\right).\end{array}$

This last formula shows how several ideas mentioned previously can be combined to form new products.

#### References

[1] Weisstein, Eric W. “Infinite Product.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/InfiniteProduct.html

[2] A, Dieckmann, Collection of Infinite Products and Series