$x_{n}$ is the unique solution of the equation $x+\cdots+x^{n}=1$ in $[0,1]$. How to find the equivalent infinitesimal of $x_{n}-1/2$

Question

$x_{n}$ is the unique solution of the equation $x+\cdots+x^{n}=1$ in $[0,1]$. It's easy to prove the limit of $x_{n}$ is $1/2$. I want to study the equivalent infinitesimal of $(x_{n}-1/2)$. But I have no idea. Any help will be thanked.

Answer 1

$$\frac{x^{n+1}-x}{x-1}=1$$ so that

$$x-\frac12=\frac{x^{n+1}}2\approx\frac1{2^{n+2}}.$$

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For better approximations write

$$\left(\left(x-\frac12\right)+\frac12\right)^{n+1}-2\left(x-\frac12\right)=0$$

and develop

$$\frac1{2^{n+1}}+\frac{n+1}{2^{n}}\left(x-\frac12\right)+\frac{n(n+1)}{2\cdot2^{n-1}}\left(x-\frac12\right)^2+\cdots-2\left(x-\frac12\right)=0.$$

For small degrees, you obtain an analytical expression of the roots.

The first approximations are

$$x-\frac12=\frac1{2^{n+2}}$$

$$x-\frac12=\frac1{2^{n+2}-n-1}$$

$$x-\frac12=\frac1{2^{n+1}-n-1+\sqrt{(2^{n+1}-n-1)^2-2n(n+1)}}$$