This is from an past paper question (France mid 80s), and I will translate in English the best I can.
$f$ is a differentiable function over $\mathbb{R}$ such that $f(0)=0$ and $f'(0)=1$
$(u_n)_{n \in \mathbb{N}}$ is a sequence such that:
- $\forall n \in \mathbb{N}, u_{n+1}=f(u_n)$
- $\forall n \in \mathbb{N}, u_{n}>0$
- $\lim_{n \to \infty} u_n =0$
Let $\alpha>1$ and $a>0$ be real numbers, and suppose $\lim_{x \to 0^+}((x-f(x))x^{-\alpha}=a$
Depending on the values of the real numbers $\alpha$ and $p>0$, determine the convergence/divergence of the series $\sum {u_n}^p$
I believe I have proven the convergence for $p \geq \alpha$:
$u_n- u_{n+1} \sim_{n \to \infty} a {u_n}^{\alpha} $, so since $\sum (u_n- u_{n+1})=u_0$, the series $\sum (a {u_n}^{\alpha})=a\sum {u_n}^{\alpha}$ converges.
It was proven in a previous question that for $p>q>0$, if $\sum {u_n}^q$ converges then $\sum {u_n}^p$ converges (and if $\sum {u_n}^p$ diverges then $\sum {u_n}^q$ diverges)
From this it follows that for $p \geq \alpha, \sum {u_n}^p$ converges.
Now for $p<\alpha$ I think the series $\sum {u_n}^p$ diverges, but I have not proven it (I have possibly a proof for $p \leq \alpha-1$ but it seems a bit complicated and it leaves open $\alpha-1<p<\alpha$)
Any suggestions?
As you have noticed, we have that $u_n-u_{n+1}$ is equivalent to $au_n^{\alpha}$ as $n\to +\infty$, and this imply that $u_n/u_{n+1}$ converge to $1$. Now put for $b>0\in \mathbb{R}$, $v_n=\frac{1}{u_{n+1}^b}-\frac{1}{u_n^b}=\frac{u_n^b-u_{n+1}^b}{u_n^bu_{n+1}^b}$. We have that $u_n^b-u_{n+1}^b=u_n^{b}(1-(\frac{u_{n+1}}{u_n})^b)$ is equivalent to $bu_n^{b}(1-\frac{u_{n+1}}{u_n})$ (use the fact that $x^b-1$ is equivalent to $b(x-1)$ if $x\to 1$), and using all that we find that $v_n$ is equivalent to $abu_n^{\alpha-b-1}$, and choosing $b=\alpha-1$, that $v_n$ has for limit $a(\alpha-1)$. Now use Cesaro's theorem, to find an explicit equivalent to $u_n$, and it is easy to finish.