Easy way to use monotone convergence theorem on $a_{n+1} = (1-\frac{1}{k})\cdot a_n + \frac{r}{k} \cdot {a_{n}}^{-k+1}$ I'm looking for an easy way to use the monotone convergence theorem on $(a_n)_{n \in \mathbb{N} \cup \{0\}}$ with
$
k ~ \in \left [ ~ 2, \infty \right ) \cap \mathbb{N} \\
r ~ \in \left [ ~ 0, \infty \right ) \\
a_0 ~ \in \left ( ~ 0, \infty \right ) \\
~ \\
a_{n+1} = (1-\frac{1}{k})\cdot a_n + \frac{r}{k} \cdot {a_{n}}^{-k+1}$
It's bounded below by $0$ because $(1-\frac{1}{k}) \geq 0$, $a_0 \geq 0$, $\frac{r}{k} \geq 0$ and ${a_{0}}^{-k+1} \geq 0$. Therefore $\forall n \in \mathbb{N} \cup \{0\} ~ : ~ a_n \geq 0$. (Easy to prove by induction.)
However, I'm still looking for an easy way to show that $(a_n)$ is increasing/decreasing.
(The series converges to $r^{\frac{1}{k}}$ by the way...)
 A: Assume WLOG $\,r=1\,$ (otherwise substitute $a_n \mapsto \sqrt[k]{r}\,a_n\,$).
First, we'll show that $f(a) = \left(1-\cfrac{1}{k}\right)a + \cfrac{1}{k\,a^{k-1}} \ge 1\,$ for all $\,\forall a \gt 0\,$, which follows from:
$$
\frac{k-1}{a^{k-1}}\,\big(f(a)-1\big) = \frac{a^k}{k}+\frac{1}{k(k-1)}-\frac{a^{k-1}}{k-1} = \frac{a^k-1}{k}-\frac{a^{k-1}-1}{k-1} \ge 0
$$
The latter inequality is due to the function $g(x)=\cfrac{a^x-1}{x}$ being monotonically increasing, which follows from the convexity of $\,a^x\,$.
Therefore $\,a_1=f(a_0) \ge 1\,$. By induction, suppose $a_n \ge 1\,$, then $a_{n+1}=f(a_n) \ge 1\,$, and:
$$\frac{a_{n+1}}{a_{n}}= 1 - \frac{1}{k} + \frac{1}{k \,a_n^{k-1}} \le 1 \;\;\implies\;\; a_{n+1} \le a_n$$
The sequence $a_n$ is therefore monotonically decreasing for $n \ge 1\,$. Since it is bounded below by $0\,$, it is also convergent, and its limit $A$ can be determined by passing the recursion to the limit: $$\require{cancel} \bcancel{A} = \left(\bcancel{1}-\cfrac{1}{k}\right)A + \cfrac{1}{k\,A^{k-1}} \;\;\implies\;\; A = 1$$
