The polynomial function is $$R_n(x)=\sum_{p=2}^{n+1}\frac{(p-1)(2n-p)!}{n!(n+1-p)!}x^p$$
It says the following:
For large $n$, the asymptotic form of $R_n(1/\beta)$ can be obtained by looking at the values of $p$ which dominate the sum
- $p$ of order $1$ for $\beta > 1/2$,
- $p$ of order $\sqrt n$ for $\beta = 1/2$, and
- $p \approx (1 - 2\beta)/(1-\beta)n$ for $\beta < 1/2$
Then how to find the asymptotic solution?
Assume $\beta = 1/x > 0$. Let $p = 2 + n u$, then $$\frac {(p - 1) (2 n - p)!} {n! (n + 1 - p)!} \beta^{-p} \sim f(u) e^{n \phi(u)}, \quad n \to \infty, \\ f(u) = \frac u {\beta^2} \sqrt {\frac {1 - u} {2 \pi n (2 - u)^3}}, \\ \phi(u) = -u \ln \beta - (1 - u) \ln(1 - u) + (2 - u) \ln(2 - u).$$ The maximum of $\phi(u)$ is located at $u_0 = (2 \beta - 1)/(\beta - 1)$, which is inside the interval $(0, 1)$ when $\beta < 1/2$. Then the sum is asymptotically equivalent to the corresponding integral and, by Laplace's method, the integral is a.e. to $$n f(u_0) \int_{\mathbb R} e^{n (\phi(u_0) + \phi''(u_0) u^2/2)} du = (1 - 2 \beta) (\beta (1 - \beta))^{-n - 1}.$$ If $\beta > 1/2$, the maximum is at the left endpoint and the sum is not a.e. to the corresponding integral. Instead, since we're interested only in small values of $p$, we can replace the summand with its approximation for fixed $p$: $$\frac {(p - 1) (2 n - p)!} {n! (n + 1 - p)!} \beta^{-p} \sim \frac {(p - 1) 2^{2 n - p}} {\beta^p \sqrt {\pi n^3}}, \quad n \to \infty.$$ The sum of the remainder is asymptotically smaller, and the original sum is a.e. to $$\sum_{p \geq 2} \frac {(p - 1) 2^{2 n - p}} {\beta^p \sqrt {\pi n^3}} = \frac {2^{2 n}} {(2 \beta - 1)^2 \sqrt {\pi n^3}}.$$