A sequence of numbers $x_1, x_2, x_3, ... ,$ starts with $x_1 = 2$, and, if you know any term $x_n$, you can find the next term $x_n+1$ using the formula $x_{n=1} = \frac{1}{2} \left( x_n + \frac{3}{x_n} \right)$.

Solution by Andaleeb of Woodhouse Sixthform College, London.

For the iteration $$x_{n+1} = \frac{1}{2} \left( x_n + \frac{3}{x_n} \right)$$ We notice that when $x_n = 1.732050808$, so is $x_{n+1}$. Squaring these terms we get $x_1^2 = 4, x_2^2 = 3.0625, ... , x_5^2 = 3$ and the rest of the other terms are the same!! This implies that when $x_n \approx \sqrt{3}$ so is $x_{n+1}$ and the values of $x_n$ tend to the limit $\sqrt{3}$. This special property can easily be proven. Assume that the limit exists, so $x_{n+1} = x_n = x$, then solve the equation $$X = \frac{1}{2}\left(X + \frac{3}{N} \right)$$ If we test it for $N = 3$, we see that $x_{29} = 1.44224957$, which is what the calculator gives for the cube root of 3. Testing it for $N = 8$, we get $x_1 = 2$, which is the right answer. By experimentation you can soon discover for yourself that it is not safe to assume that the same method works finding fourth roots using the iteration formula. $$x_{n+1} = \frac{1}{2} \left( x_n + \frac{N}{x_n^3} \right)$$ There is work to do to show that the iteration $x_{n+1} = F(x_n)$ converges to a limit $L$ if and only if $-1 < F'(L) < 1$.