Rise and Fall

Calculus Level 4

Suppose $a > b > 0$ and $a_1 =\dfrac{a+b}{2}$ , $b_1= \sqrt{ab}$ . And for $n \ge 2$ , $a_n$ and $b_n$ are defined as $\large a_n = \frac{1}{2}(a_{n-1} +b_{n-1})$ and $\large b_n = \sqrt{a_{n-1} b_{n-1}}$ respectively.

Then choose the right statement for the sequences $a_n$ and $b_n$ .

a_n

is increasing and

b_n

is increasing

a_n

is decreasing and

b_n

is decreasing

a_n

is decreasing and

b_n

is increasing

a_n

is increasing and

b_n

is decreasing

1 solution

Chew-Seong Cheong
Jul 25, 2016

Since $a$ and $b$ are positive, so are $a_n$ and $b_n$ and we can apply. AM-GM inequality. Now, we have $a_1 = \dfrac {a+b}2 > \sqrt{ab}$ , $\implies a_1 > b_1$ , since $a \ne b$ , equality cannot occur.

Now $a_2 = \dfrac {a_1+b_1}2 < \dfrac {a_1+a_1}2$ $\implies a_2 < a_1$ , similarly $a_n < a_{n-1}$ , therefore, $a_n$ is decreasing.

And $b_2 = \sqrt{a_1b_1} > \sqrt{b_1^2}$ $\implies b_2 > b_1$ , similarly $b_n > b_{n-1}$ , therefore, $b_n$ is increasing.

Clever use of similar word.

Aakash Khandelwal - 4 years, 10 months ago

Rise and Fall

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