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.
 A: 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 $$\sqrt{2}=\sum _{n=0}^{\infty } \frac{(-1)^{n-1} (2 n-3)\text{!!}}{(2 n)\text{!!}}$$
A: 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 $\sqrt{2}$. 
A: 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.
A: $$\sqrt{2}=\frac{1}{\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{6^2}\right) \left(1-\frac{1}{10^2}\right) \left(1-\frac{1}{14^2}\right) \cdots}$$
$$\sqrt{2}=\left(1+\frac{1}{1}\right) \left(1-\frac{1}{3}\right) \left(1+\frac{1}{5}\right) \left(1-\frac{1}{7}\right) \cdots$$
$$\sqrt{2}=1+\frac{1}{2}-\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots$$
$$\sqrt{2}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}$$
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
$$\sqrt{2}=2 \left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots$$
A: 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 $\sqrt2$:
$$
\begin{array}{l}
\sqrt2&=&\left(1-\frac48\right)^{-1/2}&=&\sum_{k=0}^\infty\binom{2k}{k}\frac1{8^k}\\
\sqrt2&=&\frac43\left(1-\frac4{36}\right)^{-1/2}&=&\frac43\sum_{k=0}^\infty\binom{2k}{k}\frac1{36^k}\\
\sqrt2&=&\frac75\left(1-\frac4{200}\right)^{-1/2}&=&\frac75\sum_{k=0}^\infty\binom{2k}{k}\frac1{200^k}\\
\sqrt2&=&\frac{24}{17}\left(1-\frac4{1156}\right)^{-1/2}&=&\frac{24}{17}\sum_{k=0}^\infty\binom{2k}{k}\frac1{1156^k}\\
\sqrt2&=&\frac{41}{29}\left(1-\frac4{6728}\right)^{-1/2}&=&\frac{41}{29}\sum_{k=0}^\infty\binom{2k}{k}\frac1{6728^k}\\
\sqrt2&=&\frac{140}{99}\left(1-\frac4{39204}\right)^{-1/2}&=&\frac{140}{99}\sum_{k=0}^\infty\binom{2k}{k}\frac1{39204^k}\\
\sqrt2&=&\frac{239}{169}\left(1-\frac4{228488}\right)^{-1/2}&=&\frac{239}{169}\sum_{k=0}^\infty\binom{2k}{k}\frac1{228488^k}\\
\end{array}
$$
A: A fast series that seems to produce the same result as the Babylonian method is given by
$$\sqrt{2}=\frac{3}{2}-\sum_{k=0}^\infty \frac{2\sqrt{2}}{(17+12\sqrt{2})^{2^k}-(17-12\sqrt{2})^{2^k}}$$
This question asks for a similar one starting from $\dfrac{99}{70}$.
A: Hint: You might one to consider f(x) = (x+2)^(1/2) and find the Taylor series of f about x = 0. The series you obtained gives you a series for 2^(1/2).
