IMC 2003 · Day 2 · P12

Consider the generating function $f(x) = \sum\limits_{n=0}^{\infty} a_n x^n$. By induction $0 < a_n \le 1$, thus this series is absolutely convergent for $|x| < 1$, $f(0) = 1$ and the function is positive in the interval $[0, 1)$. The goal is to compute $f\left( \frac{1}{2} \right)$.

By the recurrence formula, $$f'(x) = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{a_k}{n-k+2}\, x^n = \sum_{k=0}^{\infty} a_k x^k \sum_{n=k}^{\infty} \frac{x^{n-k}}{n-k+2} = f(x) \sum_{m=0}^{\infty} \frac{x^m}{m+2}.$$ Then $$\begin{align*} \ln f(x) &= \ln f(x) - \ln f(0) = \int_0^x \frac{f'}{f} = \sum_{m=0}^{\infty} \frac{x^{m+1}}{(m+1)(m+2)} = \\ &= \sum_{m=0}^{\infty} \left( \frac{x^{m+1}}{(m+1)} - \frac{x^{m+1}}{(m+2)} \right) = 1 + \left( 1 - \frac{1}{x} \right) \sum_{m=0}^{\infty} \frac{x^{m+1}}{(m+1)} = 1 + \left( 1 - \frac{1}{x} \right) \ln \frac{1}{1-x}, \end{align*}$$ $$\ln f \left( \frac{1}{2} \right) = 1 - \ln 2,$$ and thus $f\left( \frac{1}{2} \right) = \dfrac{e}{2}$.