Finding the global minimum of $x^{n}+x^{n-1}+...+1$ for even $n.$ Inspired by this question, I am curious if there is an asymptotic of the global minimum of the function: $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1.$$
In the referred question, I showed that $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1 =x^{2n-2}(x+\dfrac{1}{2})^2+\dfrac{3}{4}x^{2n-4}(x+\dfrac{2}{3})^2+\dfrac{4}{6}x^{2n-6}(x+\dfrac{3}{4})^2+\dots+ \dfrac{n+1}{2n}(x+\dfrac{n}{n+1})^2+\dfrac{n+2}{2n+2}> \dfrac{n+2}{2n+2}.$$ 
 Since the method was completely elementary, I figured this was a loose bound. However, it turned out to be surprisingly close to the actual values for when $n=2,3$ and sharp $n=1.$ 
When $n=2$, the minimum is $0.673753\approx\dfrac{2}{3}=0.66$ and  when $n=3$, the minimum is $0.635\approx \dfrac{5}{8} = 0.625.$
Thus, is it possible to obtain an asymptotic for $\min_{x\in\mathbb{R}}f_n(x)$, as $n\to\infty?$ 
A closed form solution would be even better, but that just seem hopeless. 
 A: Claim. The minimum of $f_n$ tends to $\frac12$ when $n \to \infty$.
Let $a_n$ denote the minimum of $f_n(x) = \frac{x^{2n+1}-1}{x-1}$ for $x \in \mathbb{R}$. 
For fixed $t>0$ and $n > t$ we also have $$a_n \leq f_n\left(-1 + \frac{t}n\right) = \frac{-1 + (-1 + \frac{t}{n})^{1 + 2 n}}{-2 + \frac{t}{n}} = \frac{1 + (1-\frac{t}{n})^{2n+1}}{2-\frac{t}n} \leq \frac{1+ (1-\frac{t}{n}) e^{-2t}}{2-\frac{t}{n}},$$
because $(1-\frac{t}{n})^n \leq e^{-t}$. This means that for any $t>0$ we have
$$
\frac{1 + \frac{2}{n}}{2+\frac{2}{n}} \leq a_n \leq \frac{1+ (1-\frac{t}{n}) e^{-2t}}{2-\frac{t}{n}}
$$ 
for sufficiently large $n$. For $n \to \infty$ the lower bound tends to $\frac12$, whereas the upper bound tends to $\frac{1+e^{-2t}}{2}$. Because this holds for any $t$, we find $\lim_{n \to \infty} a_n = \frac12$. 
A: 
Claim. Let $a_n=\min_{x\in\mathbb{R}}f_n(x)$, then $\lim_{n\to \infty}a_{n}=1/2$.

We first note that
$f_{n}(x)=\sum_{k=1}^n(x^{2k-2}(x(1+x))+1\geq 1$ for $x\not \in [-1,0]$. Therefore, since $f_n(0)=1$,
$$a_n=\min_{x\in\mathbb{R}}f_n(x)=\min_{x\in[-1,0]}\frac{1-x^{2n+1}}{1-x}=\min_{t\in[0,1]} \frac{1+t^{2n+1}}{1+t}.$$
Now for $t\in [0,1]$,
$$\frac{1+t^{2n+1}}{1+t}\geq \frac{1}{1+t}\geq \frac{1}{2}\implies a_n\geq\frac{1}{2}\implies \lim_{n\to \infty}a_{n}\geq \frac{1}{2}.$$
On the other hand, for $t_n=1-\frac{1}{\sqrt{n}}\in [0,1]$
$$a_n\leq \frac{1+{t_n}^{2n+1}}{1+t_n}\implies \lim_{n\to \infty}a_{n}\leq \lim_{n\to \infty}\frac{1+{t_n}^{2n+1}}{1+t_n}=\frac{1}{2}$$
and the claim is proved.
