Prove that $a_n\to 0$ but $\sum a_n = \infty$.

Question

Let $f:[-1,1]\to\mathbb{R}$ be a $C^2-$function with $f(0)=0$, $f^{\prime}(0)=1$ and $f^{\prime}(x)>1$ for all $x>0$. Let $a_1\in (0,1)$ and define recursively $a_{n+1}$ by equation $f(a_{n+1})=a_n$. Prove that $a_n\to 0$ but $\sum a_n = \infty$.

I was doing this exercise but I am kind of confused by the definition of the sequence, we should have the injection for this and I cannot say anything about the function around $-1$. If I consider $a_n \in (0,1)$ it works well and I can prove that $a_n\to 0$ , with the respective conditions, but I also know that can exist negative points $a_i$ near $0$ for some $i$ so it becomes a problem. And what about $\sum a_n = \infty$? I am out of ideas for this.

Answer 1

As Martin R says, it’s quite necessary to assume $0 < a_n < 1$, and then $a_n$ decreases to zero. Behavior of the series $a_n$ is (of course) trickier.

We know by Taylor that $a_n=f(a_{n+1})=a_{n+1}+\int_0^{a_{n+1}}{tf’’(a_{n+1}(1-t))dt}=a_{n+1}+a_{n+1}^2(c +o(1))$ where $c=f’’(0)/2$. Thus $a_{n+1}^{-1}-a_n^{-1}=a_{n+1}^{-1}(1-(1+(c+o(1))a_{n+1})^{-1})=\frac{(c+o(1))a_{n+1}}{a_{n+1}(1+(c+o(1))a_{n+1})}=c+o(1)$ (so $c \geq 0$). It follows that $0<a_n^{-1}=O(n)$ so $a_n \geq C/n$ for some positive constant $C$.