Convergence of $u_{n+1} = \frac{u_n^2}{2}+\frac{1}{2}$

Question

I am looking for the convergence of $u_n$, defined by

$$u_{n+1} = \frac{u_n^2}{2}+\frac{1}{2}, \qquad u_0=a$$

I found that $u_n$ is an increasing sequence and it converges to $\infty$. I feel that I missed something.

Answer 1

$$u_{n+1} - u_n = \frac12 (u_n^2 - 2u_n + 1) = \frac12(u_n-1)^2.$$ So if $a=1$ then $u_n = 1$ for all $n>0$. If $a=-1$, then $u_1=1$ and then it stays at $1$. Otherwise, the sequence is always increasing, as you say.

If $|a|<1$, then inductively, as $u_{n+1}$ is the average of $u_n^2 < 1$ and $1$, we also have $u_{n+1}\le 1$. Any possible limit $L$ solves $$ 2L=L^2+1\iff (L-1)^2 = 0 \iff L = 1$$ so $u_n\to 1$.

If $|a|>1$ then as it must increase, but cannot increase to $1$, it therefore increases without bound.

Answer 2

For $a>1$ it is true that $u_n$ diverges to $\infty$ but monotonicity doesn’t suffices to conclude that. Moreover what about all the other cases $a\le1$?