# Math: convergence questions

1. Investigate the convergence of the sequence defined recursively $bya(1) =\sqrt{2}$

$a(n+1) = \sqrt{2*a(n)}$

2. Suppose a, b are positive real numbers. Prove that $2ab/(a+b) <= \sqrt{ab}$

1. Prove that $a(n) \leq 2$ by induction.

$a(1) = \sqrt{2} \leq 2$ true

suppose $a(n-1) \leq 2$

then $a(n) = \sqrt{2*a(n)}=\sqrt{2}*\sqrt{a(n)} \leq \sqrt{2}*\sqrt{2} = 2$ true

Prove that $a(n)$ is increasing.

$a(n+1)/a(n) = \sqrt{2*a(n)} /a(n) = \sqrt{2/a(n)} \geq \sqrt{2/2} = 1$

hence $a(n+1) \geq a(n)$ thus it is increasing.

Since a(n) is upper bounded by 2 and it is increasing it is convergent having the upper limit 2

2.

$2ab/(a+b) \geq \sqrt{ab}$ (1)

since $\sqrt{a*b}$ is positive and a and b are positive we can divide (simplify) both terms by $\sqrt{ab}$

$2*\sqrt{ab} /(a+b) \leq 1$ (2)

$(a+b) \geq 2\sqrt{(ab)}$ (3)

$a-2\sqrt{(ab)} +b \geq 0$ (4)

$(\sqrt{(a)} -\sqrt{(b)})^2 \geq 0$ (5)

The last equality is obvious since a and b are real positive numbers and a square of a number is always greater than or equal to 0.