I asked this question of MSE, but to no avail; alas, here I am.  

Let $k>0$, $C\geq 1$, $\alpha \in (0,1]$, and let $(x\_n)\_{n\geq 1}$, be a sequence of real numbers given by the recursion
$$
x\_{n+1} = k |C|^{\alpha} + |x\_{n}|^{\alpha} \qquad x\_0=0.
$$
Is there a simple "non-recursive" expression for $x\_{n}$, for $n>0$? At the very worst, is there a tight-upper bound for $x\_n$ only depending on $C,k,$ and $\alpha$? I was trying to obtain an upper-bound by the geometric sums $\sum\_{i=1}^n{k C}^{\alpha}$ but I'm no longer convinced this is correct.



Closed-form for recursive "geometric-like" recursion
It seems extremely unlikely that a simple "non-recursive" expression for $x\_n$ is possible. However, let us obtain an exact upper bound on the $x\_n$'s.


Let $a:=\alpha\in(0,1]$ and $b:=k|C|^a\in(0,\infty)$. It is clear that $x\_n\ge0$ for all $n$. So,
$$x\_0=0,\quad x\_1=b,\tag{1}$$
and
$$x\_{n+1}=g(x\_n)\tag{2}$$
for $n\ge1$, where $g(u):=b+u^a$. To avoid trivialities, assume that $a\ne1$, so that $a\in(0,1)$. Then $h(u):=g(u)-u$ is concave in real $u\ge0$, with $h(0)=b>0$ and $h(\infty-)=-\infty$. So, there is a unique root $u\_b\in(0,\infty)$ of the equation $h(u)=0$ and, moreover, $h\ge0$ on the interval $[0,u\_b]$, that is,
$$g(u)\ge u\quad \forall u\in[0,u\_b],\tag{2.5}$$
whereas
$$b+u\_b^a=g(u\_b)=u\_b.\tag{2.75}$$
In particular, it follows that
$$u\_b\ge b.\tag{3}$$


Let us show, by induction on $n\ge0$, that
$$x\_n\le u\_b\tag{4}$$
for all $n\ge0$. Indeed, by (1) and (3), (4) holds for $n=0,1$. Supposing now that (4) holds for some $n\ge1$, we have
$$x\_{n+1}=g(x\_n)\le g(u\_b)=u\_b,$$
in view of (2) and because the function $g$ is increasing. So, (4) holds for all $n\ge0$.


Now (2), (2.5), and (4) imply
$$x\_{n+1}=g(x\_n)\ge x\_n.$$
So, $(x\_n)$ is a nondecreasing sequence in $[0,u\_b]$. Therefore and in view of (2) and (2.75),
$$x\_n\uparrow u\_b$$
as $n\uparrow\infty$.


Thus, in view of (4), $u\_b$ is the best upper bound on the $x\_n$'s.




---


If $a$ is rational, then the unique root $u\_b$ of the equation $u\_b=g(u\_b)$ will be algebraic in $b$. E.g., if $a=1/2$, then $u\_b=(1+2 b+\sqrt{1+4 b})/2$.


