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?


1 Answer 1


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).

  • $\begingroup$ I'm a bit lost; how did you get the limit for $c_n$ and $P'(c_n)$? $\endgroup$ Commented Feb 26 at 5:26
  • 1
    $\begingroup$ $c_n\in ]1/2;x_n[$ 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 $\lim P'(c_n)=\lim 1/(1-c_n)^2=4$ $\endgroup$ Commented Feb 26 at 10:10

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