, 2006

The proof goes by induction. For $n=1$ the formula yields $a_{1}=1 / 2$. Take $n \geq 1$, assume $a_{1}, \ldots, a_{n}>0$ and write the recurrence formula for $n$ and $n+1$, respectively as
$$\sum_{k=0}^{n} \frac{a_{k}}{n-k+1}=0 \quad \text{ and } \quad \sum_{k=0}^{n+1} \frac{a_{k}}{n-k+2}=0 .$$
Subtraction yields
$$\begin{aligned} 0=(n+2) \sum_{k=0}^{n+1} & \frac{a_{k}}{n-k+2}-(n+1) \sum_{k=0}^{n} \frac{a_{k}}{n-k+1} \\ & =(n+2) a_{n+1}+\sum_{k=0}^{n}\left(\frac{n+2}{n-k+2}-\frac{n+1}{n-k+1}\right) a_{k} \end{aligned}$$
The coefficient of $a_{0}$ vanishes, so
$$a_{n+1}=\frac{1}{n+2} \sum_{k=1}^{n}\left(\frac{n+1}{n-k+1}-\frac{n+2}{n-k+2}\right) a_{k}=\frac{1}{n+2} \sum_{k=1}^{n} \frac{k}{(n-k+1)(n-k+2)} a_{k} .$$
The coefficients of $a_{1}, \ldots, a_{n}$ are all positive. Therefore, $a_{1}, \ldots, a_{n}>0$ implies $a_{n+1}>0$.