Finding the global minimum of $x^{n}+x^{n-1}+...+1$ for even $n.$

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 $\lim_{n \to \infty} a_n = \frac12$.