Let $f(x)=\frac13(x^2+x+1)$. Define $f_0(x)=x$ and for all $k\ge 1$, $f_k(x)=f(f_{k-1}(x))$. Let $a_k=f_k(0)-f_{k-1}(0)$. Is $\sum\limits_{k=1}^{\infty} ka_k$ infinte?
I found out the followings:
$a_k= f_{k}(0)-f_{k-1}(0)=\frac13(1-f_{k-1}(0))^2\ge 0$
$0\le f_k(0) \le 1$ and $\lim\limits_{k\to \infty} f_k(0) =1$
I tried ratio test but $\lim\limits_{k\to\infty} \frac{a_{k}}{a_{k-1}}=1$, so couldnot proceed any further. Any suggestions or hints?
So here is the answer:
Note that $1-f(x)=1-x-\frac13(1-x)^2$ $$\implies \begin{aligned}\frac1{1-f(x)} =\frac{1}{(1-x)(1-\frac13(1-x))} & =\frac1{1-x}[1+\frac13(1-x)+O((1-x)^2)]\\ & =\frac1{1-x}+\frac13+O(1-x)\end{aligned}$$ Therefore $$\frac1{1-f_{k+1}(0)}=\frac1{1-f_k(0)}+\frac13+O(1-f_k(0))$$ Therefore adding the above equalities for each $k$ we get $$\frac1{1-f_n(0)}=1+\frac{n}{3}+\sum_{i=0}^{n-1} O(1-f_i(0))$$ Now note that RHS is $O(n)$, hence $1-f_n(0)$ is $O(\frac1n)$. Hence $\displaystyle \sum_{i=1}^{n-1} O(1-f_i(0))$ is $O(\log n)$. Hence $$\lim_{n\to \infty} \frac{n(1-f_n(0))}{3}=1$$ This implies $1-f_n(0) \sim \frac3n$ Then $a_k =f_k(0)-f_{k-1}(0) \sim \frac{3}{k^2}$. Hence $\sum ka_k$ diverges!