Compare Roots

Notice that $x^n - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + \cdots + x + 1)$ and also that $x^n - 202x + 201 = (x - 1)(x^{n - 1} + x^{n - 2} + \cdots + x - 201).$ Put $s_n = x^{n - 1} + x^{n - 2} + \cdots + x + 1$ and write the equation as $200(x - 1)s_n + n(x - 1)(s_n - 202) = 0 \implies (x - 1)[(n + 200)s_n - 202n] = 0.$ The roots of this equation are $x = 1$ along with any roots of $s_n = \frac {202n}{n + 200}. \tag{1}$ Observe that $s_n(0) = 1$ and $s_n(1) = n$ , and when $n > 2$ , $n - \frac {202n}{n + 200} = \frac {n^2 - 2n + 200} {n + 200} > 0 \implies n > \frac {202n}{n + 200}.$ This implies that $s_n(0) < \frac {202n}{n + 200} < s_n(1) \tag{2}$ and thus that $(1)$ has a real root $x_n$ in $(0,1)$ . Also, $s_n' > 0$ for all $x \in (0,1)$ , so $s_n$ is increasing on $(0,1)$ . This makes $x_n$ the unique root of $(1)$ in $(0,1)$ , and it is the root asserted in the problem.

What remains is to show that $x_{28} \approx 0.99087$ is the minimal value of $x_n$ . Put $a = 109/110 \approx 0.99091$ and notice that $s_n(a)$ is the partial sum of a geometric sequence with ratio $a$ , so $s_n(a) = \frac {1 - a^n}{1 - a}.$ Consider the limits $\lim_{n \rightarrow \infty} s_n(a) = \lim_{n \rightarrow \infty} \frac {1 - a^n}{1 - a} = \frac 1{1 - a} = 110 \quad \text{and} \quad \lim_{n \rightarrow \infty} \frac {202n}{n + 200} = 202.$ The limits show that, for sufficiently large $n$ , $s_n(a) < \frac {202n}{n + 200} < s(1). \tag{3}$ By the same reasoning applied to $(2)$ in the previous paragraph, $x_n \in (a,1)$ when $n$ is sufficiently large. Therefore, $x_n < a$ can be true for only finitely many values of $n$ . In fact, $x_n < a$ only when $21 \leq n \leq 37$ , and for these $n$ -values, $x_n$ attains its minimum value at $n = \boxed{28}$ .

More can be said about the $x_n$ . For $n \geq 28$ , as $n$ increases, $x_n$ increases as well, but $x_n$ does not approach $1$ . This is because, if $n > 3$ and $a \geq 1 - 202^{-1}$ , then $s_n(a) = \frac {1 - a^n}{1 - a} > \frac {202n}{n + 200}.$ In other words, $(3)$ does not hold when $n > 3$ and $a \geq 1 - 202^{-1}$ , so $x_n \not \in [a,1)$ and thus $x_n < 1 - 202^{-1}$ . But $(3)$ does hold when $n = 3$ and $a = 1 - 202^{-1}$ , and $x_3 \approx 0.99507 > 1 - 202^{-1}$ . Therefore, $x_3$ is the greatest value of $x_n$ , and all the $x_n$ occur in a very narrow interval: for $n \geq 3$ , $x_n \in [x_{28},x_3] \approx [0.99087,0.99507].$