Show that the sequence $\{a_n\}$ is divergent?

Question

Let, $a_1=2$ and for $n\ge 2$, $a_{n+1}=a_n^2-a_n+1$. Prove that the sequence $\{a_n\}$ is divergent.

It is clear that , $a_{n+1}-a_n=(a_n-1)^2\ge 0$ which implies $a_{n+1}\ge a_n.$ So the sequence $\{a_n\}$ is monotonic increasing. So it is sufficient to show that $\{a_n\}$ is unbounded above !

If $\{a_n\}$ is bounded then then $\{a_n\}$ is convergent, say converges to $l$. i.e., $a_n \to l$ as $n\to \infty$. Then, from the given relation we have, $l=l^2-l+1$ , which implies $l=1$, contradicts that $a_1=2$. So we are done !

Any other way to show that the given sequence is unbounded above ?

Answer 1

Assume $a_k\geq 2$. Then $$ a_{k+1} = a_k^2-a_k + 1 = a_k(a_k-1) + 1\geq a_k\cdot 1 + 1= a_k + 1 $$ which shows that the sequence increases by at least $1$ for each step. By induction, we easily get $a_n\geq n+1$, but it could also be used to directly show non-convergence by the definition of limit (take $\epsilon = \frac12$, and go through the motions).

Answer 2

Note that $a_{n+1}-a_n=(a_n-1)^2\geqslant1$. Since $a_1=2$, this show that $a_n\geqslant n+1$.

Answer 3

A quantitative approach:

$$a_n^2-a_n+1=\left(a_n-\frac12\right)^2+\frac34$$

so that with $b_n:=a_n-\dfrac12$,

$$b_{n+1}=b_n^2+\frac14>b_n^2.$$

Hence if $b_1>1$,

$$b_n>b_1^{2^{n-1}},$$ which is an extremely fast, unbounded growth.

---

As the term $\dfrac14$ quickly becomes negligble, we can approximate the sequence with

$$a_n\approx c^{2^{n-1}}+\frac12$$ and obtain $c$ from some $a_n$.

We find

$$c\approx\sqrt[32]{a_6-\frac12}=\sqrt[32]{3263442.5}=1.597910\cdots$$