Recurrence relation - Show that there is $A$ s.t. $a_n < A \forall n \in N$

Question

Let $a_n$ be defined such that: $a_1 = 1, a_n = \frac{1}{20} a^2_{n-1} +3 \forall n>= 2 $

Show that there is $A$ s.t. $a_n < A \forall n \in N$

Thanks in advance!

Answer 1

$$a_n = \frac{1}{20} (a_{n-1})^2 +3$$

$$a_1=1$$

Let $L \in \mathbb{R}$ with $L>1$ satisfy,

$$L=\frac{1}{20}L^2+3$$

Take $L=10-2\sqrt{10}$

Suppose there exists $N$ such that $1<a_N < L$. Then because the function $x \mapsto x^2$ is increasing for $x>0$ we must have,

$$1^2 <a_{N}^2 < L^2$$

$$\frac{1}{20}1^2 <\frac{1}{20}a_{N}^2 < \frac{1}{20}L^2$$

$$\frac{1}{20}1^2+3<\frac{1}{20}a_{N}^2+3 < \frac{1}{20}L^2+3=L$$

$$1<\frac{61}{20}< a_{N+1} < L$$

So because $L>a_2>1$ it follows by the above that $L>a_3>1$. Reusing the above argument it then follows $L>a_4>1$. Hence by induction,

$$L>a_n>1$$

For any integer $n \geq 2$.

Answer 2

Find $A$ so that $A>1$ and $\dfrac{A^2}{20}+3 \leq A$ and we'll be done by induction.

[Hint: There is an integer $A$ which satisfies the inequality.]