Let $a_1 = 1$, and let $a_{n+1} = \left(a_n\right)^2 + 1$. What is the growth rate of $a_n$? Even better, is there a closed-form formula for $a_n$?
Since each step is "basically squared", I'd expect it to grow something like $2^{2^{n-1}}$. In fact, I'm pretty sure we have $2^{2^{n-2}} \leq a_n \leq 2^{2^{n-1}}$.
Do either of the ratios
$$ \frac{a_n}{2^{2^{n-2}}} $$
or
$$ \frac{a_n}{2^{2^{n-1}}} $$
converge to a nonzero value as $n \rightarrow \infty$?
For the sequence $b_n$ constructed in the comments, it is not difficult to see that $0\le b_n\le 1/2$ (We can use induction to prove the second inequality and the first one is obvious).
Then, by Bolzano-Weierstrass theorem there is a subsequence which converges. Let $\{b_{n_k}\}$ be such a subsequence. Let the limit be denoted by $b$. Then the recursion relation shows that $b$ should satisfy $b^2=b\implies b=0,1$. Since $b_n\le 1/2,\ b=0$.
Choose $\epsilon>0$ such that $\epsilon<1/2$. Then, $\exists K, N$ such that $b_{n_k}<\epsilon\ \forall k\ge K$ and $c_n<\epsilon/2\ \forall n\ge N$. Let $N_1=\max(n_K,N)$ and let $K_1$ be such that $n_{K_1}\ge N_1$. Then, $b_{n_{K_1}+1}<\epsilon^2+\epsilon/2<\epsilon$ since $\epsilon<1/2$. Applying this logic further to $n\ge n_{K_1}+1$ shows that $b_{n}<\epsilon\ \forall n\ge n_{K_1}$. Hence $\forall \epsilon<1/2$, $\exists N$ such that $b_{n}<\epsilon\ \forall n\ge N$. Since this is true for all $0<\epsilon<1/2$, $\lim_{n\to \infty}b_n=0$.