Convergence of the sequence $\displaystyle a_{n+1} = \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}})$ So, I'm trying to establish the convergence of the sequence $\{a_n\}$ defined as:
$$a_{n+1} = \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}}) \text{  , } n\in \mathbb{N} $$
where $a>0$ and $a_1>0$.
So, before you vote to close this question because it has been probably asked numerous times earlier, I want to say that I've already established the convergence on my own, but I have reached two conclusions that can contradict each other and I can't understand why this happens. Maybe I'm just too tired now, but I've been thinking over this for almost an hour and I have no more energy to solve the issue on my own:
The first result is obtained just by applying AM-GM:
$$a_{n+1} = \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{(p-1)}}) \geq \sqrt[p]{a_n^{p-1} \cdot \frac{a}{a_n^{p-1}}} = \sqrt[p]{a}$$
So, for $n \in \mathbb{N}$, $a_{n+1} \geq \sqrt[p]{a}$
So, this makes sense to me, but one can easily see that we also have:
$$a_{n+1}-a_{n}= \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}}) - \frac{1}{p}( (p-1)a_{n-1}+\frac{a}{a_{n-1}^{p-1}})$$
$$\implies p(a_{n+1}-a_{n})= (p-1)(a_{n}-a_{n-1}) + a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$$
So, this is where my confusion starts:
Let's prove by induction that if $a_1 < \sqrt[p]{a}$ then $\{a_n\}$ is increasing and if $a_1 > \sqrt[p]{a}$ then $\{a_n\}$ is decreasing. If $a_1 = \sqrt[p]{a}$ then $\{a_n\}=\{a_1\}$ becomes a constant sequence.
So, to establish the base of induction, if we manipulate the equations we get:
$$a_2 - a_1 = \frac{a-a_1^p}{pa_1^{p-1}}$$
This proves the base of induction in all the three cases. The rest follows immediately from $\displaystyle p(a_{n+1}-a_{n})= (p-1)(a_{n}-a_{n-1}) + a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$
Now, here is the contradiction:
If $a_1 < \sqrt[p]{a}$ then $\{a_n\}$ must be an increasing sequence as I just proved by induction, while, the AM-GM inequality says that all the terms of the sequence beyond $a_2$ must be greater than or equal to $\sqrt[p]{a}$! This implies that the sequence must diverge to $+\infty$ I guess. Right?
But I'm starting to think, by doing numerical examples, that in a real situation, if we start with $a_1 < \sqrt[p]{a}$, the sequence first jumps to some number above $\sqrt[p]{a}$ and then it starts to decrease until it converges to $\sqrt[p]{a}$ at the end.
Can someone tell me what's going on?

EDIT
OK, I had a sign error as gammatester pointed out. I modified my proof and now I think I have solved the problem correctly:
$$a_{n+1}-a_{n} = \frac{a-a_{n+1}^p}{pa_{n+1}^{p-1}}$$
Since $a_{n+1} \geq \sqrt[p]{a}$ we conclude $a_{n+1} \leq a_{n}$. So, the sequence is decreasing and bounded below, hence, it's convergent.

What you showed is exactly this:

If $a_n\lt\sqrt[p]{a}$ then $a_{n+1}\gt a_n$. 

To use this at rank $n+1$ to deduce that $a_{n+2}\gt a_{n+1}$, one would need to know that $a_{n+1}\lt\sqrt[p]{a}$, which is not so. Actually, $(p-1)x+a/x^{p-1}\geqslant p\sqrt[p]{a}$ for every $x\gt0$ hence $a_n\geqslant\sqrt[p]{a}$ for every $n\geqslant2$. So, no contradiction of ZF here...
