Upper Bound on $\frac{1+ax^{n+1}}{1+ax^n}$ for $x \in (0,1)$

Question

I am looking for an upper bound on $$\frac{1+ax^{n+1}}{1+ax^n}$$ in the regime $x \in (0,1)$ where $a>0$ and $n$ is a non-negative integer.

I would like it to be of the form \begin{align} \frac{1+ax^{n+1}}{1+ax^n} \le f(a,x,n) \end{align} where $f(a,x,n)$ is a polynomial. Moreover, I would like to be `good' around $x=1$.

Clearly, there is a trivial bound \begin{align} \frac{1+ax^{n+1}}{1+ax^n} \le 1. \end{align}

I was thinking that we can do a Tailor expansion around the minimum of this function.

Answer 1

We have the following polynomial upper bound: \begin{align} \frac{1+ax^{n+1}}{1+ax^n} &= \frac{1+ax^n - ax^n + ax^{n+1}}{1+ax^n} \\ &= 1 - \frac{ax^n(1-x)}{1+ax^n}\\ &\le 1 - \frac{ax^n}{(1+a)^2}(1+2a - ax^n)(1-x) \end{align} where we have used $$\frac{1}{1+ax^n} = \frac{1}{1+a}\cdot\frac{1}{1 - \frac{a(1-x^n)}{1+a}} \ge \frac{1}{1+a}\left(1 + \frac{a(1-x^n)}{1+a}\right).$$

Answer 2

The trivial bound is the best you can do, because $$\lim_{x \to 1^-} \frac{1+ax^{n+1}}{1+ax^n} = 1.$$

Answer 3

If we Taylor expand the function around $x=1$, we have $$\frac{1+a\,x^{n+1}}{1+a\,x^n}=1+t+\frac{n }{a}t^2+O\left(t^3\right)\qquad \text{where} \qquad t=-\frac{a (1-x)}{a+1}$$

Hoping that it could help !