series approximations of irrational numbers

binomial expansions
To approximate square roots, find a ratio of squares a^2/b^2 close to the number n, then use \sqrt{n}=\frac{a}{b}(\frac{nb^2}{a^2})^\fr{2}, as \frac{nb^2}{a^2} will be close to 1.

    \begin{align*} \sqrt{2}&=2^{\fr{2}}=(\frac{49}{25})^\fr{2}\cdot (\frac{50}{49})^\fr{2} \\ &= \frac{7}{5}\cdot(\frac{49}{50})^{-\fr{2}} = \frac{7}{5}\cdot(1-\fr{50})^{-\fr{2}}\\ &=\frac{7}{5}\Big[ 1-\binom{-\fr{2}}{1}\fr{50}+\binom{-\fr{2}}{2}\fr{50^2}-\binom{-\fr{2}}{3}\fr{50^3}+ \dots\Big] \\ &= \frac{7}{5}\Big[ 1+\frac{1}{2\cdot 50}+\frac{1\cdot 3}{2\cdot2\cdot2\cdot 50^2}+\frac{1\cdot 3\cdot5}{2\cdot3\cdot2\cdot2\cdot2\cdot 50^3}+ \cdots\Big] \\ &= \frac{7}{5} (1+\fr{100}+\frac{3}{20000}+\frac{1}{400000}+\cdots) \\ &\approx \frac{7}{5} \times 1.0101525 \\ &= 1.4142135 \\ &(\sqrt{2}=1.41421356237\dots) \end{align*}

    \begin{align*} \sqrt{5}&=\frac{9}{4}\sqrt{\frac{5\times16}{81}}=\frac{9}{4}(1+\fr{80})^{-\fr{2}} \\ &=\frac{9}{4}\Big[1-\frac{1}{2\cdot80}+\frac{3}{2^2\cdot2\cdot80^2}-\frac{3\cdot5}{2^3\cdot2\cdot3\cdot80^3}\dots \Big] \\ &=\frac{9}{4}(1-\fr{160}+\frac{3}{51200}-\fr{1638400}\dots) \\ &\approx\frac{9}{4}(1-\frac{2029}{327680}) \\ &=2.236067962646484375 \\ &(\sqrt{5}=2.236067977\dots \end{align*}