Starting with $a_1=\frac{1}{2}(ab)b_1=\sqrt[\phantom{2}]{ab}n=2,3,4,cdots{a}_n=\frac{1}{2}(a_{n-1}{b}_{n-1})$ and $b_n=\sqrt[\phantom{2}]{a_{n-1}{b}_{n-1}}{a}_{n-1}>a_n>b_n>b_{n-1}n=2,3,4,$cdots$\underset{n}{lim}{a}_n=\underset{n}{lim}{b}_{n}0,$ form the arithmetic mean $a_1=\frac{1}{2}(ab)$ and the geometric mean $b_1=\sqrt[\phantom{2}]{ab}.$ For $n=2,3,4,$cdots, set

$a_n=\frac{1}{2}(a_{n-1}{b}_{n-1})$ and $b_n=\sqrt[\phantom{2}]{a_{n-1}{b}_{n-1}}$

$(a)$ Show that

$a_{n-1}>a_n>b_n>b_{n-1}$

for $n=2,3,4,$cdots

Hint: you may need $to$ use math induction.

$(b)$ Show that the two sequences converge and $\underset{n}{lim}{a}_n=\underset{n}{lim}{b}_n.$