How to compute a lot of digits of $\sqrt{2}$ manually and quickly? After having read the answers to calculating $\pi$ manually, I realised that the two fast methods (Ramanujan and Gauss–Legendre) used $\sqrt{2}$. So, I wondered how to calculate $\sqrt{2}$ manually in an accurate fashion (i.e., how to approximate its value easily).
 A: you can use the formula 
$$\frac{a_{n+1}}{b_{n+1}}=\frac{a_n^2+2b_n^2}{2a_nb_n}$$
if we take an initial value of $\sqrt{2}$ as $\frac{3}{2}$ 
now the new value will become
$$\frac{a}{b}=\frac{3^2+2*2^2}{2*2*3}=\frac{17}{12}$$
the new value of $a=17$ and $b=12$
and then continue
A: One really easy way of approximating square roots surprisingly accurately was actually developed by the Babylonians. 
First they made a guess at the square root of a number $N$--let this guess be denoted by $r_1$. Noting that
$$
r_1\cdot\left(\frac{N}{r_1}\right)=N,
$$
they concluded that the actual square root must be somewhere between $r_1$ and $N/r_1$. Thus, their next guess for the square root, $r_2$, was the average of these two numbers:
$$
r_2 = \frac{1}{2}\left(r_1+\frac{N}{r_1}\right).
$$
Continuing in this way, in general, once we have reached the $n$th approximation to the square root of $N$, we find the $(n+1)$st using
$$
r_{n+1}=\frac{1}{2}\left(r_n+\frac{N}{r_n}\right).
$$
All that you really need to do is make a moderately decent guess of the square root of a number and then apply this method two or three times and you should have quite a good approximation. 
For $\sqrt{2}$, simply using a guess of $1$ and applying this method three times (the algebra involved is remarkably simple) yields an approximation of
$$
\frac{577}{408}\approx \color{red}{1.41421}\color{blue}{568627},
$$
whereas
$$
\sqrt{2}\approx \color{red}{1.41421}\color{green}{356237}.
$$
That's quite a good approximation using an easy and quick manual method. 
A: Another technique might be to use the Taylor series
$$(1+x)^{1/2} = 1+ \frac 12 x - \frac 18 x^2 + \frac{1}{16} x^3 - \frac{5}{128} x^4 +\cdots.$$
The coefficients of this series are 
$\frac{(-1)^k }{k!} \left(\frac12\right) \left(-\frac12\right)\left(-\frac32\right)\cdots\left(\frac32 - k\right)$.
You can plug in $x=1$ so that the series evaluates to $\sqrt2$, 
but the series converges faster if you start with a rational approximation
of $\sqrt2$ and use the Taylor series to compute a correction factor,
for example $\sqrt 2 = 1.4 \cdot \left(1 + \frac{1}{49}\right)^{1/2}.$
