Problem #PRU-61297

Problems Calculus Number sequences Limit of a sequence, convergence Algebra Sequences Recurrent relations Recurrent relations (other)

15–16 3.0

Problem

The Babylonian algorithm for deducing $\sqrt{2}$ . The sequence of numbers ${x_{n}}$ is given by the following conditions: $x_{1} = 1$ , $x_{n + 1} = \frac{1}{2} (x_{n} + 2 / x_{n})$ ( $n \geq 1$ ).

Prove that $lim_{n \to \infty} x_{n} = \sqrt{2}$ .

To see the solution register and get verified.