Asymptotic behavior of the real root of $kx^{k+1} -(k+2)x^{k} + 1$

Question

I am interested in the asymptotic behavior of the largest real root of the following polynomial: $$ \mathcal{P}_k \colon \quad k x^{k+1} - (k+2) x^k + 1$$

Let $\alpha_k$ denote the largest real root of the polynomial $\mathcal{P}_k$. It is easy to check that $\lim_{k \rightarrow \infty} \alpha_k = 1$. However, I would like to understand the rate of convergence of the limit. More precisely, I want to either compute the limit exactly, $$ \lim_{k \rightarrow \infty} \alpha_k^k,$$ or find some interval (the smallest possible) so that $\lim_{k \rightarrow \infty} \alpha_k^k \in [a, b]$. I have no idea whether a similar question was already studied and what tools I could use to tackle this question. From some numerical experiments, I get that $\lim_{k \rightarrow \infty} \alpha_k^k \in [6.2, 7]$ but can't prove it rigorously.

How I got there: I was analyzing a non-linear problem. After computing its Lagrange multiplier relaxation, I ended up with the polynomial $\mathcal{P}_k$ where $\alpha_k^k$ is the value I am trying to maximize.

Any help is appreciated.

Answer 1

Note that $$ P_k(x)=x^k(kx-k-2)+1$$ and $$\begin{align} P_k'(x)&=kx^{k-1}(kx-k-2)+kx^k\\&=kx^{k-1}((k+1)x-(k+2)),\\ \end{align}$$ so that $P_k$ can have local extrema at most at $x=0$ and $x=1+\frac1{k+1}$. From $P_k(0)=1$, $P_k(1)=-1$, $P_k(1+\frac2k)=1$, the last critical point must be a minimum and in particular $P_k(1+\frac1{k+1})<-1$. We conclude $$ 1+\frac1{k+1}<\alpha_k<1+\frac2k.$$ This give us $$\tag1e\approx \left(1+\frac1{k+1}\right)^k<\alpha_k^k<\left(1+\frac 2{k}\right)^k\approx e^2.$$ If we define $x_k\lesssim y_k$ as "For all $\epsilon>0$, for almost all $k$, $x_k<y_k+\epsilon$", then this can be written as $$ e\lesssim\alpha_k^k\lesssim e^2.$$ Note that $$\tag2\alpha_k^k=\frac1{2-k(\alpha_k-1)}.$$ We can make this last step more precise: Suppose we have $a,b$ such that $$a\lesssim\alpha_k^k\lesssim b. $$ Then from $(2)$, we conclude $$1+ \frac{2-\frac1{b}}k\lesssim \alpha_k\lesssim 1+ \frac{2-\frac1{a}}k$$ and so $$ e^{2-\frac1b}\lesssim \alpha_k^k\lesssim e^{2-\frac1a}.$$ Starting with $a=e$ and $b=e^2$, we obtain quite small intervals after a few iterations. The intervals converge to a point $y$ with $e^{2-\frac1y}=y$.

Answer 2

Write $\ln{\alpha_k}=\beta_k$. Then $e^{k\beta_k}(k+2-ke^{\beta_k})=1$. So $1 \leq \alpha_k \leq 1+2/k$. Thus $e^{\beta_k} = 1+\beta_k+r_k$, with $k^2r_k \leq C$. Therefore, taking logarithms, one finds $k\beta_k+\ln(2-k\beta_k-kr_k)=0$. In particular, $k\beta_k+\ln(2-k\beta_k)> 0$, so it follows $k\beta_k < 1.9$.

As $kr_k \rightarrow 0$, it implies that $\frac{2-k\beta_k-kr_k}{2-k\beta_k} \rightarrow 1$. Therefore $k\beta_k+\ln(2-k\beta_k) \rightarrow 0$. Thus, the limit $\ell$ of $\alpha_k^k=e^{k\beta_k}$ satisfies the equation $\ell(2-\ln{\ell})=1$, as Aleksei Kulikov's comment tells.