Suppose $a,b\in\mathbb{N}$ and, moreoever, $1\leqslant a$ and $b\geqslant a+2$.
I am considering the polynomial $$ f_{a,b}(x):=x^{2b}-\frac{x^{b-a}-1}{x-1} $$ which has exactly one positive (simple) root $x_{a,b}$ and, moreover, $x_{a,b}>1$. In particular, $\lim_{a\to\infty}x_{a,b}=1$.
I am trying to analyse at which rate $x_{a,b}$ tends to $1$ as $a\to\infty$. To this end, I make the ansatz $$ x_{a,b}=1+y_{a,b} $$ and now try to analyse at which rate $y_{a,b}\to 0$ as $a\to\infty$, say, $y_{a,b}=\frac{1}{a}+o(1/a)$ or whatever the correct rate might be.
Do you have any idea how to get this?
My first attempt was to plug the ansatz for $x_{a,b}$ in the polynomial: \begin{align*} &(1+y_{a,b})^{2b}-\frac{(1+y_{a,b})^{b-a}-1}{(1+y_{a,b})-1}=0\\ &\Leftrightarrow (1+y_{a,b})^{2b+1}-(1+y_{a,b})^{2b}-(1+y_{a,b})^{b-a}+1=0 \end{align*}
Using the binomial theorem, I wrote the last equation as \begin{align*} &\sum_{k=1}^{2b+1}\binom{2b+1}{k}y_{a,b}^k-\sum_{k=1}^{2b}\binom{2b}{k}y_{a,b}^k-\sum_{k=1}^{b-a}\binom{b-a}{k}y_{a,b}^k\\ &=y_{a,b}(1-(b-a))+\sum_{k=2}^{2b+1}\binom{2b+1}{k}y_{a,b}^k-\sum_{k=2}^{2b}\binom{2b}{k}y_{a,b}^k-\sum_{k=2}^{b-a}\binom{b-a}{k}y_{a,b}^k\\ &=0 \end{align*}
Factoring out $y_{a,b}$, what I get is \begin{equation*} y_{a,b}\cdot \left(1-b+a+\sum_{k=1}^{2b}\binom{2b+1}{k+1}y_{a,b}^k-\sum_{k=1}^{2b-1}\binom{2b}{k+1}y_{a,b}^k-\sum_{k=1}^{b-a-1}\binom{b-a}{k+1}y_{a,b}^k\right)=0. \end{equation*}
This equation is fulfilled exactly if $y_{a,b}=0$ (what seems not to be helpful for my purpose) or if \begin{equation} \sum_{k=1}^{2b}\binom{2b+1}{k+1}y_{a,b}^k-\sum_{k=1}^{2b-1}\binom{2b}{k+1}y_{a,b}^k-\sum_{k=1}^{b-a-1}\binom{b-a}{k+1}y_{a,b}^k=b-a-1. \end{equation}
Maybe this last equation can help to get the desired Information about $y_{a,b}$; however, I don’t see how.
I am thankful for your ideas.
For brevity, I will denote $n = 2b$, $k = b-a$ and drop the index $a,b$ in $y := y_{a,b}$. Our equation is $$(1+y)^n = \frac{(1+y)^k - 1}{y}$$ which we can rewrite as $$(1+y)^k = 1 + y (1+y)^n$$ and taking logarithms, we get $$k \ln(1+y) = \ln(1 + y (1+y)^n)$$ Now, assuming $y (1+y)^n \underset{a \to \infty}{\longrightarrow} 0$, we get the equivalents $$k y \underset{a \to \infty}{\sim} y (1+y)^n$$ dividing by $y$ and taking logarithms yields $$\ln(k) \underset{a \to \infty}{\sim} n \ln(1+y) \underset{a \to \infty}{\sim} n y$$ and finally we get $$y \underset{a \to \infty}{\sim} \frac{\ln(k)}{n} = \frac{\ln(b-a)}{2b}$$
EDIT : There was a mistake initially in assuming the condition $y (1+y)^n \underset{a \to \infty}{\longrightarrow} 0$ was automatic, but it's only true under the assumption that $k \ln(k) = o(n)$. In this section, we prove the following three properties are equivalent :
1) $y (1+y)^n \underset{a \to \infty}{\longrightarrow} 0$
2) $y \sim \frac{\ln(k)}{n}$
3) $k \ln(k) = o(n)$
We have just proved that $1) \Longrightarrow 2)$.
Let's prove $2) \Longrightarrow 3)$. We assume 2), that is $y \sim \frac{\ln(k)}{n}$. We can then compute that \begin{eqnarray*} y (1+y)^n &=& \frac{\ln(k)}{n} e^{n \ln(1 + \frac{\ln(k)}{n})} \\ &=& \frac{\ln(k)}{n} e^{\ln(k) - \frac{(\ln(k))^2}{2n} + o\left(\frac{(\ln(k))^2}{n}\right)} \\ &=& \frac{\ln(k)}{n} k \, e^{- \frac{(\ln(k))^2}{2n} + o\left(\frac{(\ln(k))^2}{n}\right)} \\ &\sim& \frac{k \ln(k)}{n} \end{eqnarray*} Which implies that $\frac{k \ln(k)}{n} \to 0$, which is property 3).
And finally, we prove that $3) \Longrightarrow 1)$. We assume $k \ln(k) = o(n)$. Let $\epsilon > 0$ be a fixed parameter. We prove that for $a$ sufficiently large $$\frac{\ln(k) - \epsilon}{n} < y < \frac{\ln(k) + \epsilon}{n}$$ To do that, we compute the asymptotic expansion of $f_{a,b}\left(1+ \frac{\ln(k) \pm \epsilon}{n}\right)$ and look at its sign. Indeed we have \begin{eqnarray*} \left(1 + \frac{\ln(k) \pm \epsilon}{n}\right)^n &=& e^{n \ln\left(1+ \frac{\ln(k) \pm \epsilon}{n}\right)}\\ &=& e^{(\ln(k) \pm \epsilon) - \frac{(\ln(k) \pm \epsilon)^2}{2n} + o\left(\frac{(\ln k \pm \epsilon)^2}{2n}\right)}\\ &=& k \, e^{\pm\epsilon} e^{- \frac{(\ln(k) \pm \epsilon)^2}{2n} + o\left(\frac{(\ln k \pm \epsilon)^2}{2n}\right)}\\ &=& k \, e^{\pm\epsilon} + o(k) \end{eqnarray*} And \begin{eqnarray*} \frac{\left(1 + \frac{\ln(k) \pm \epsilon}{n}\right)^k-1}{\frac{\ln(k) \pm \epsilon}{n}} &=& \frac{n}{\ln(k) \pm \epsilon} \left(e^{k \ln\left(1+ \frac{\ln(k) \pm \epsilon}{n}\right)}-1\right) \\ &=& \frac{n}{\ln(k) \pm \epsilon} \left(e^{\frac{k(\ln(k) \pm \epsilon)}{n} + o\left(\frac{k(\ln(k) \pm \epsilon)}{n}\right)} - 1 \right)\\ &=& k + o(k) \end{eqnarray*} So we finally get $$f_{a,b}\left(1+ \frac{\ln(k) \pm \epsilon}{n}\right) \sim (e^{\pm\epsilon}-1) \, k $$ This means that for $a$ sufficiently large, we have $f_{a,b}\left(1+ \frac{\ln(k) - \epsilon}{n}\right) < 0$ and $f_{a,b}\left(1+ \frac{\ln(k) + \epsilon}{n}\right) > 0$ and so there is a root of $f_{a,b}$ between those two boundaries. But because there is only one positive root, it must be $y$, and we deduce $$\frac{\ln(k) - \epsilon}{n} < y < \frac{\ln(k) + \epsilon}{n}$$
And now we can asymptotically bound the quantity $y (1+y)^n$ by
$$ \underbrace{\frac{\ln(k) - \epsilon}{n} \left(1 + \frac{\ln(k) - \epsilon}{n}\right)^n}_{\sim \frac{k (\ln(k)-\epsilon) e^{-\epsilon}}{n}} < y (1+y)^n < \underbrace{\frac{\ln(k) + \epsilon}{n} \left(1 + \frac{\ln(k) + \epsilon}{n}\right)^n }_{\sim \frac{k (\ln(k)+\epsilon) e^{\epsilon}}{n}}$$
where both bounds converge to $0$ from the assumption that $k \ln(k) = o(n)$. Which proves 1).