Asymptotic Solution of Recurrence Relations I have a recurrence relation,
\begin{align}
0 \leq A_{n+1} \leq A_n - c_1 {A_n}^{m} + \frac{c_2}n, ~~~\forall n\geq 1\label{rec}\tag{1}
\end{align}
where $c_1$ and $c_2$ are positive constants, and $m\in \mathbb{N}.$ 
I want to show that, $A_n = O(\frac 1{n^\epsilon})$ for a positive constant $\epsilon$.
My approach : 
Let us look at the continuous differential equation corresponding to this recurrence.
$$A_{n+1} - A_n \leq -c_1 {A_n}^m + \frac{c_2}n$$
$$\approx \frac{\partial y(x)}{ \partial x} = -c_1 y(x)^m + \frac{c_2}x $$
We can solve this differential equation for $m=2$. And show, using induction, that,
$$0 \leq A_n \leq y(n), ~\forall n\geq 1.$$
The induction proof follows from the following conditions,


*

*$y(n) = O\left(\frac1{n^\epsilon}\right)$

*$y(1) \leq \left(\frac1{m c_1} \right)^{\frac1{m-1}} $

*$y(n) - c_1 y(n)^m + \frac{c_2}n \leq y(n+1), ~\forall n\geq 1$
But since, we cannot solve the differential equation for higher $m$ (I am not aware of a general approach for that), this approach seems improbable to lead to a solution for general values of $m$.
Are there any other methods to approach this problem? Any references would be appreciated!
 A: Let $$F(n) \triangleq \left(\frac{\kappa_2 + \alpha}{\kappa_1 ~n}\right)^{1/m},$$ where $\alpha$ is the smallest positive value such that, $\frac{\alpha}n \geq {\left(\frac{\kappa_2+\alpha}{\kappa_1}\right)^{1/m}} \left(\left(\frac1n\right)^{1/m}-\left(\frac1n\right)^{1/m}\right), ~\forall n\geq1$. Then, we can prove using induction that for a constant integers $N_0, n_0$, 
    \begin{equation}\label{lemma4_res}
  A_{n} \in [0,F(n-N_0)],~\forall n\geq n_0
 \end{equation}
To this affect, first observe that $A_n \stackrel n{\rightarrow} 0$.
Thus, there must exist integers $N_0, n_0$ such that $n_0\geq N_0$, and $A_{n_0} \leq F(n_0-N_0) \leq \left(\frac1{\kappa_1 m}\right)^{1/m}$. Then, to prove the induction we only need to show, $A_{n+1} \in [0,F(n+1-N_0)]$ or equivalently 
$$F(n-N_0) - \kappa_1 F^m(n-N_0) + \frac{\kappa_2}n \leq F(n+1-N_0).$$
The LHS in above satisfies, 
    \begin{align*}
  F(n-N_0) -  F(n+1-N_0) &= \left(\frac{\kappa_2+\alpha}{\kappa_1}\right)^{1/m} \left(\left(\frac1{n-N_0}\right)^{1/m}-\left(\frac1{n+1-N_0}\right)^{1/m} \right)\\
  &\leq \frac{\alpha}{n-N_0}.
 \end{align*}
    from the condition on $\alpha$.   And the RHS satisfies, 
    \begin{align*}
  \kappa_1 F^m(n-N_0) - \frac{\kappa_2}n &= -\frac{\kappa_2}n + \frac{\kappa_2+\alpha}{n-N_0} \\
  &\geq \frac{\alpha}{n-N_0}.
 \end{align*}
