Square Root of 2 - Series and Product Representations

Series and Product Representations

The identity, along with the infinite product representations for the sine and cosine, leads to products such as

$\frac{1}{\sqrt 2} = \prod_{k=0}^\infty \left(1-\frac{1}{(4k+2)^2}\right) = \left(1-\frac{1}{4}\right) \left(1-\frac{1}{36}\right) \left(1-\frac{1}{100}\right) \cdots$

and

$\sqrt{2} = \prod_{k=0}^\infty \frac{(4k+2)^2}{(4k+1)(4k+3)} = \left(\frac{2 \cdot 2}{1 \cdot 3}\right) \left(\frac{6 \cdot 6}{5 \cdot 7}\right) \left(\frac{10 \cdot 10}{9 \cdot 11}\right) \left(\frac{14 \cdot 14}{13 \cdot 15}\right) \cdots$

or equivalently,

$\sqrt{2} = \prod_{k=0}^\infty \left(1+\frac{1}{4k+1}\right) \left(1-\frac{1}{4k+3}\right) = \left(1+\frac{1}{1}\right) \left(1-\frac{1}{3}\right) \left(1+\frac{1}{5}\right) \left(1-\frac{1}{7}\right) \cdots.$

The number can also be expressed by taking the Taylor series of a trigonometric function. For example, the series for gives

The Taylor series of with and using the double factorial gives

$\sqrt{2} = \sum_{k=0}^\infty (-1)^{k+1} \frac{(2k-3)!!}{(2k)!!} = 1 + \frac{1}{2} - \frac{1}{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} - \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots.$

The convergence of this series can be accelerated with an Euler transform, producing

$\sqrt{2} = \sum_{k=0}^\infty \frac{(2k+1)!}{(k!)^2 2^{3k+1}} = \frac{1}{2} +\frac{3}{8} + \frac{15}{64} + \frac{35}{256} + \frac{315}{4096} + \frac{693}{16384} + \cdots.$