<s>[INST] Infinite series for $ \sqrt 2 $ What is infinite series for $ \sqrt 2 $? I don't mean continued fraction. That kind of series such as like for $e, \pi, $etc.
 [/INST] $$\sqrt{2}=\frac{1}{\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{6^2}\right) \left(1-\frac{1}{10^2}\right) \left(1-\frac{1}{14^2}\right) \cdots}$$
$$\sqrt{2}=\left(1+\frac{1}{1}\right) \left(1-\frac{1}{3}\right) \left(1+\frac{1}{5}\right) \left(1-\frac{1}{7}\right) \cdots$$
$$\sqrt{2}=1+\frac{1}{2}-\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots$$
$$\sqrt{2}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}$$
https://en.wikipedia.org/wiki/Square_root_of_2#Series_and_product_representations
The first one is the true answer. It is in the format of the natural number $e = \left(1+\frac{1}{\infty}\right)^\infty$ except it's a minus sign instead and a reciprocal. Consider it's reduced form here:
$$\sqrt{2}=\frac{1}{\left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots}$$
This is the most reduced true form of the infinite series that is $\sqrt2$. An amazing property of $\sqrt2$ is that the reciprocal is equal to exactly $\frac{1}{2}$ of its value. So
$$\sqrt{2}=2 \left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots$$
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