# Proof by induction

• May 13th 2010, 08:02 PM
Mathman87
Proof by induction
Let a1=3 and a(n+1) =($\displaystyle {a}^{2}n+10$)/7 for every n$\displaystyle \geq$1

Prove by induction that 2
$\displaystyle \leq$an$\displaystyle \leq$5 for every n$\displaystyle \geq$1

Deduce that an+1$\displaystyle \leq$an for every n$\displaystyle \geq$1

Show the sequence {an} converges and determine its limit.
• May 13th 2010, 09:03 PM
simplependulum
Quote:

Originally Posted by Mathman87
Let a1=3 and a(n+1) =($\displaystyle {a}^{2}n+10$)/7 for every n$\displaystyle \geq$1

Prove by induction that 2$\displaystyle \leq$an$\displaystyle \leq$5 for every n$\displaystyle \geq$1

Deduce that an+1$\displaystyle \leq$an for every n$\displaystyle \geq$1

Show the sequence {an} converges and determine its limit.

Induction on $\displaystyle n$ , $\displaystyle a_1 =3 \in[2,5]$ and assume it is true that $\displaystyle 2 \leq a_n \leq 5$ and we have

$\displaystyle a_{n+1} = \frac{a_n^2 + 10}{7} \leq \frac{ 5^2 + 10}{7} = \frac{35}{7} = 5$

Also ,

$\displaystyle a_{n+1} = \frac{a_n^2 + 10}{7} \geq \frac{2^2 + 10}{7} = \frac{14}{7} = 2$

Therefore , $\displaystyle 2\leq a_{n+1} \leq 5$

Hence we deduce it is true that $\displaystyle 2\leq a_n \leq 5 ~~\forall n \in \mathbb{N}$

Consider $\displaystyle a_{n+1} - a_n = \frac{a_n^2 + 10}{7} -a_n = \frac{a_n^2 - 7a_n + 10 }{7} = \frac{(a_n - 2)(a_n -5 )}{7}$

$\displaystyle a_n - 2 \geq 0$ because $\displaystyle a_n \geq 2$

but $\displaystyle a_n - 5 \leq 0$ from $\displaystyle a_n \leq 5$

Therefore , $\displaystyle a_{n+1} - a_n$ is a non-positive number or $\displaystyle a_{n+1} - a_n \leq 0 ~,~ a_{n+1} \leq a_n$

Note that the sequence is bounded and it is monotonic decreasing , use some theorems to show that the limit exists and find it .