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).

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. 
