Infinite series for $ \sqrt 2 $

Solution 1:

The generating function for the Central Binomial Coefficients is $$ (1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag{1} $$ We can plug $x=\frac18$ into $(1)$ to get $$ \begin{align} \sqrt2 &=\sum_{k=0}^\infty\binom{2k}{k}\frac1{8^k}\\ &=\sum_{k=0}^\infty\frac{(2k-1)!!}{4^kk!}\tag{2} \end{align} $$

Alternatively, we could plug $x=-\frac14$ into $(1)$ and double the result to get $$ \begin{align} \sqrt2 &=2\sum_{k=0}^\infty\binom{2k}{k}\left(-\frac14\right)^k\\ &=2\sum_{k=0}^\infty(-1)^k\frac{(2k-1)!!}{(2k)!!}\tag{3} \end{align} $$ However, the error in the partial sum of $(3)$ is $O\left(\frac1{\sqrt{k}}\right)$. The error in the partial sum of $(2)$ is $O\left(\frac1{2^k\sqrt{k}}\right)$, which yields much faster convergence.

Using Continued Fractions, we get rational approximations to $\sqrt2$ that can be used with $(1)$ to get other series for $\sqrt2$: $$ \begin{array}{l} \sqrt2&=&\left(1-\frac48\right)^{-1/2}&=&\sum_{k=0}^\infty\binom{2k}{k}\frac1{8^k}\\ \sqrt2&=&\frac43\left(1-\frac4{36}\right)^{-1/2}&=&\frac43\sum_{k=0}^\infty\binom{2k}{k}\frac1{36^k}\\ \sqrt2&=&\frac75\left(1-\frac4{200}\right)^{-1/2}&=&\frac75\sum_{k=0}^\infty\binom{2k}{k}\frac1{200^k}\\ \sqrt2&=&\frac{24}{17}\left(1-\frac4{1156}\right)^{-1/2}&=&\frac{24}{17}\sum_{k=0}^\infty\binom{2k}{k}\frac1{1156^k}\\ \sqrt2&=&\frac{41}{29}\left(1-\frac4{6728}\right)^{-1/2}&=&\frac{41}{29}\sum_{k=0}^\infty\binom{2k}{k}\frac1{6728^k}\\ \sqrt2&=&\frac{140}{99}\left(1-\frac4{39204}\right)^{-1/2}&=&\frac{140}{99}\sum_{k=0}^\infty\binom{2k}{k}\frac1{39204^k}\\ \sqrt2&=&\frac{239}{169}\left(1-\frac4{228488}\right)^{-1/2}&=&\frac{239}{169}\sum_{k=0}^\infty\binom{2k}{k}\frac1{228488^k}\\ \end{array} $$

Solution 2:

As suggested by NovaDenizen, Taylor expansion of $f(x) = \sqrt{x + 1}$ has a general term which write $$\frac{(-1)^{n-1} (2 n-3)\text{!!} x^n}{(2 n)\text{!!}}$$ Setting $x=1$ then leads to $$\sqrt{2}=\sum _{n=0}^{\infty } \frac{(-1)^{n-1} (2 n-3)\text{!!}}{(2 n)\text{!!}}$$

Solution 3:

It is too easy to give series with irrational terms. So let us try for rational. One can note that $\sqrt{2}\approx 1.41421356\dots$. Thus an infinite series for $\sqrt{2}$ is $$1+\frac{4}{10}+\frac{1}{10^2}+\frac{4}{10^3}+\frac{2}{10^4}+\frac{1}{10^5}+\frac{3}{10^6}+\frac{5}{10^7}+\frac{6}{10^8}+\cdots.$$ The only issue is with the $\cdots$. We have not given an explicit expression for the $n$-th term.

If we use the Maclaurin series for $(1-x)^{-1/2}$, evaluated at $x=1/2$, we can get an explicit series with rational terms that converges to $\sqrt{2}$.

Solution 4:

As indicated in the other answers, you use the binomial series for $\sqrt{1+x}$. However, $x=1$ is at the boundary of the region of convergence, so you first reduce the problem algebraically by observing that, as robjohn has used in his answer, $\sqrt2=(\frac12)^{-1/2}=(1-\frac12)^{-1/2}$ or with even smaller offsets as

$$\sqrt{2}=\frac32\sqrt{\frac89}=\frac32\sqrt{1-\frac19}=\frac32\left(1+\frac18\right)^{-\frac12}$$

or

$$\sqrt{2}=\frac75\sqrt{\frac{50}{49}}=\frac75\sqrt{1+\frac1{49}}=\frac75\left(1-\frac1{50}\right)^{-\frac12}$$

With these smaller values for $x$ under the root in any of those 4 expressions, convergence of the binomial series is much more rapid.