Prove that the sequence $a_n = a_{n-1}^2 + a_1$ when $a_1\gt\frac{1}{4}$ has no upper bound

Question

Given the sequence $a_n = a_{n-1}^2 + a_1$ for $n > 1$ with $a_1 \gt \frac{1}{4}$ how can one show by definition that this sequence has no upper bound?

Given any sequence $\{a_n\}$, we say that the sequence is not bounded above if $\forall M \in \mathbb{R}, \exists n \in \mathbb{N}$ such that $a_n > M.$

I managed to prove that $\{a_n\}$ is a strictly monotone increasing sequence, so intuitively I thought that whatever $M$ we take, I can find a big enough $n$ that $a_n > M$.

Answer 1

Let $a_1\gt \frac{1}{4}$ and $a_n=a_{n-1}^2+a_1$ for $n>1$. Let's assume the $\lim_{n\rightarrow\infty}=a$ exists. Then taking the limit on both sides of the equation yields $a=a^2+a_1$. So we get $a^2-a+a_1=0$. This is a polynomial and has a real solution iff $1-4a_1\geq 0$, which is not the case in our case. So we know the limit doesn't exist. But as you stated the sequence is monotone increasing. So this means the sequence is not bounded from above, as if it would be, it would be convergent.

Answer 2

Define $b_n=a_n-\frac{1}{2}$ then $a_n=b_n+\frac{1}{2}$.

This gives $$b_n+\frac{1}{2}=(b_{n-1}+\frac{1}{2})^2+b_1+\frac{1}{2}\\ b_n=b_{n-1}^2+b_{n-1}+\frac{1}{4}+b_1\\ $$

Now, $a_1 >\frac{1}{4} \Leftrightarrow b_1+\frac{1}{4}>0$. Denote by $c = b_1+\frac{1}{4}>0$.

Then you have $$b_n=b_{n-1}^2+b_{n-1}+c\geq b_{n-1}+c \,.$$

Prove now by induction that $$b_n \geq (n-1)c+b_1$$

Answer 3

For $n \ge 2$, using the fact that $a_n$ is monotonically increasing,

\begin{eqnarray} a_{n+1} - a_n &=& (a_n^2 + a_1)- (a_{n-1}^2 + a_1)\\ &=& ((a_{n-1}+a_1)^2 + a_1)- (a_{n-1}^2 + a_1)\\ &=& 2a_{n-1}a_1 + a_1^2\\ &\ge& 2a_1^2 + a_1^2\\ &>& \frac{3}{16} > \frac18 \end{eqnarray}

From this, we can see that $a_n$ must be unbounded since we are adding at least $\frac18$ each time.

In particular, if $N \in \Bbb N$, then $a_{8n} > N$.

Answer 4

$a_n-a_{n-1}=a_{n-1}^2+a_{1}-a_{n-1}>(a_{n-1}-1/2)^2$ so if we suppose $a_{n}$ is bounded then $a_{n}->1/2$ but then $a_{n+1}>1/2$...

Answer 5

Suppose the sequence had an upper bound. Then it'd have a least upper bound. Let $A = \sup a_n$ and, of course, $A \ge a_1 > 0 $ and $A - a_1 \ge 0$.

Then for all $a_{n+1} \le A$ we have $a_{n+1} = a_n^2 + a_1 \le A$ so $ a_n \le \sqrt{A - a_1}$. So $\sqrt{A- a_1}$ is also an upper bound.

But $A$ was the least upper bound so $A \le \sqrt{A - a_1}$

So $A^2 \le A - a_1$ and $A^2 - A \le -a_1$ and...

$0 \le (A - 1/2)^2 = A^2 - A + 1/4 \le -a_1 + 1/4 < -a_1 a_1 = 0$

which is impossible.

(Note: we may assume $A$ and $A - a_1$ are non-negative so $a_n^2 \le A - a_1 \implies a_n \le \sqrt{A-a_1}$ and $A \le \sqrt{A- a_1} \implies A^2 \le \sqrt{A-a_1}$ etc.)