And its solution:
1 + a_n = 1 + 1/(2 + a_(n-1))
however, in the limit n -> Infinity,
we have a_n = a_(n-1). Lets define it to be 'a'.
So, a = 1/(2 + a), that is a^2 + 2 a - 1 = 0
which gives a = +/- sqrt(2) - 1
thus 1 + a = +/- sqrt(2), but the solution with
the minus sign should be rejected as the expression is
obviously positive, so we obtain sqrt(2).
