$ (a_{n})_{n\in \mathbb{N}} $ , $ a_{0}=-1 $ , $a_{n+1}=2+\int_{a_{n}}^{1} e^{-x^2}dx$

Question

Consider the sequence : $ (a_{n})_{n\in \mathbb{N}} $ , $ a_{0}=-1 $ , $a_{n+1}=2+\int_{a_{n}}^{1} e^{-x^2}dx$ , $ n = 0,1,....$

Which of the following statement is true?

$a)$ $ (a_{n+1}-a_{n})(a_{n}-a_{n-1}) \leq 0 $ , $ \forall n\in \mathbb{N}^* $

$b)$ $ a_{n} \geq 2 $ , $ \forall n\in \mathbb{N}^* $

$c)$ $ a_{n} \leq 2 $ , $ \forall n\in \mathbb{N}^* $

$d)$ $ (a_{n})_{n\in \mathbb{N}} $ is ascending

$e)$ $ (a_{n})_{n\in \mathbb{N}} $ is decreasing

As a high-scooler it's quite weird to approach this, because I can't actually integrate $e^{-x^2}$

The correct answer should be $a)$

Answer 1

Don't worry about integrating $e^{-x^2}$; that's a famous problem beyond what you need here. All you need is this fact: since the integrand is positive, the integral is positive when the lower limit is less than the upper limit, or negative if the reverse is true.

Clearly $a_1>2>1$ and $a_2<2$, ruling out b and c. Then $a_1>a_0$ and $a_2<a_1$, ruling out d and e.

If you want to prove a, I'll leave that as an exercise.

Answer 2

To solve these type of problems, my approach is eliminating incorrect choices: To do that, first, we need some estimations or approximate values of first few terms of $(a_n)_{n\in\Bbb{N}}.$

Note that $a_{1}=2+\int_{-1}^{1} e^{-x^2}dx$ and minimum value of $e^{-x^2}$ in $[-1,1]$ is $e^{-1}.$ Therefore by the trapezoidal rule, $\int_{-1}^{1} e^{-x^2}dx\gt1+e^{-1}\approx1.37.$

Now $a_0=-1$ and $a_1\gt3.37.$ Due to those values we can observe that $a_2\approx1.86\lt a_1.$
This leave out the answers, except $(a),$ which you can prove easily following my footprints :)

Answer 3

For $n = 1, 2,3,\ldots$ $$ a_{n+1} - a_n = \int_{a_n}^1 e^{-x^2}dx- \int_{a_{n-1}}^1 e^{-x^2}dx = \int_{a_n}^{a_{n-1}} e^{-x^2}dx $$ so that $(a_{n+1} - a_n)$ and $(a_{n} - a_{n-1})$ are both zero or have opposite sign. It follows that $$ (a_{n+1}-a_{n})(a_{n}-a_{n-1}) \leq 0 \, . $$.