# Limit of a sequence of the real root of a family of polynomials

Consider a sequence of polynomials $$P(x)=x^n+x^{n-1}+x^{n-2}+...+x^2+x-1, n>2$$.

(i) Prove that it has a unique positive real root $$x_n$$

(ii) Find $$\lim_{n \to ∞} 2^n (x_n - 1/2)$$

The first part was relatively simple; upon differentiating $$P(x)$$, it is easy to see that for all $$x>0$$, the function is strictly increasing, so it can at most have one root. Secondly, $$P(0)=-1$$ and $$P(1)=n-1$$ i.e. the root must lie in $$(0,1)$$. The third observation is that, as we apply the limit of n tending to infinity on the polynomial, $$x_n$$ tends to $$1/2$$ (it just becomes an infinite geometric series) which explains why the limit is an indeterminate form of $$∞.0$$

In the concise solution of the book, it was written that $$x_n = \dfrac{1}{2}+\dfrac{\theta_n}{2^{n+1}}$$ where $$\theta_n$$ lies in $$[0,1]$$ and left at that. Firstly, this step seems quite unmotivated to me and I also couldn't use it to solve the limit. Secondly, I failed to prove this equality as well (it resembled the form of LMVT so that was my starting point but it led nowhere).

Does anyone have a more intuitively well-motivated answer for the second part of this problem or any observation that I missed?

By the mean value theorem there exists $$c_n\in]\frac{1}{2};x_n[$$ such that $$P(x_n)-P(1/2)=(x_n-1/2)P'(c_n)$$. We have $$\displaystyle P'(x)=\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2}$$ and $$\lim_{n\rightarrow +\infty}c_n=1/2$$ and $$P(1/2)=-\frac{1}{2^n}$$ so $$\lim_{n\rightarrow +\infty}P'(c_n)=4$$ and $$\displaystyle x_n-\frac{1}{2}\sim \frac{1}{4}\times \frac{1}{2^n}$$ which gives you the limit (ii).
• I'm a bit lost; how did you get the limit for $c_n$ and $P'(c_n)$? Commented Feb 26 at 5:26
• $c_n\in ]1/2;x_n[$ and the limit of $x_n$ is $1/2$ hence also that of $c_n$. Then using for instance $|c_n|<2/3$ from a given rank you have that $(n+1)c_n^n$ and $nc_n^{n+1}$ have a limit which is zero and finally $\lim P'(c_n)=\lim 1/(1-c_n)^2=4$ Commented Feb 26 at 10:10