TITLE: Computing as many digits as possible of $\sqrt{2}$ with a pen and a paper in 5 minutes
QUESTION [6 upvotes]: You have to compute as many digits as possible of $\sqrt{2}$ with a pen and a paper (an eraser if you're lucky...) in 5 minutes.
What will you do? What is your justification for doing it?
The question background is algorithm efficiency and ease to perform associated computations manually.

REPLY [3 votes]: Consider the binomial series
$$\begin{align*}
(1-x)^{\frac12}&=1-\frac12x-\frac{\frac12\frac12}{2!}x^2-\frac{\frac12\frac12\frac32}{3!}x^3+\cdots\\
&=1-\frac x2\left(1+\frac12\frac x2\left(1+\frac33\frac x2\left(1+\frac54\frac x2(1+\cdots)\right)\right)\right)
\end{align*}$$
Substituting $x=\frac1{50}$ now gives a series which gives roughly 2 decimal digits per term (initially at least), and is really fast to compute by hand:
$$\begin{array}{rl}
1-\frac7{10}\sqrt2=&0.0100000000000\\
+&\phantom{0.00}00500000000\\
+&\phantom{0.0000}005000000\\
+&\phantom{0.000000}0062500\\
+&\phantom{0.00000000}00875\\
+&\phantom{0.0000000000}013125\\
=&0.0100505063388(3)
\end{array}$$
The error is easy to bound as a geometric progression with ratio $\frac1{50}$, ie. at most $\frac1{49}$ of the final added term.
Thus
$$\begin{array}{rl}
7\sqrt2&=9.899494936612(3)\\
\sqrt2&=1.414213562373(1)
\end{array}$$
which is accurate to the last (12th) digit.
The fact that this method is so well-adapted to computation by hand is largely a coincidence, because $\frac75$ is a convergent in the continued fraction of $\sqrt2$, giving us $7^2\approx\frac{100}2$. I have yet to find a variation that gives, say, $\sqrt3$ to 12 digits with a comparable amount of work.