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

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.

• How do you know that $a_n\in[-1,1]$? If it isn't, what is $f(a_n)$? – Don Thousand Feb 23 at 16:25
• From the given conditions, $f(x)>x$ for $x>0$, hence $a_n$ should be divergent (or at least sooner or later leave $[-1,1]$. – Hagen von Eitzen Feb 23 at 16:26
• This has been asked (and answered) recently: math.stackexchange.com/q/4035021/42969. – Martin R Feb 23 at 16:26
• @HagenvonEitzen: Note that the (somewhat unusual) recursion is $f(a_{n+1})=a_n$, so $(a_n)$ is decreasing. – Martin R Feb 23 at 16:27
• If $a_n \in (0,1)$ is not given then the equation $f(a_{n+1})=a_n$ may have multiple solutions for $a_{n+1}$ and the sequence is not well-defined. – Martin R Feb 23 at 16:36

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 $$0 so $$a_n \geq C/n$$ for some positive constant $$C$$.