Tenemos que $a_1-\frac{1}{2}=0\implies a_1=\frac{1}{2}$. Ahora vamos a demostrar que si $a_i>0$ $\forall 1\leq i\leq n$ entonces $a_{n+1}>0$.
Tenemos que $\sum_{k=0}^{n+1} \frac{a_{n+1-k}}{k+1}=0=a_{n+1}+\sum_{k=1}^{n+1} \frac{a_{n+1-k}}{k+1}$, entonces debemos demostrar que $\sum_{k=1}^{n+1} \frac{a_{n+1-k}}{k+1}<0$
Como $\sum_{k=0}^{n} \frac{a_{n-k}}{k+1}=\sum_{k=1}^{n+1} \frac{a_{n+1-k}}{k}=0$, tenemos que $\sum_{k=1}^{n+1} \frac{a_{n+1-k}}{k+1}=\sum_{k=1}^{n+1} a_{n+1-k}(\frac{1}{k+1}-\frac{1}{k})=\sum_{k=1}^{n+1} \frac{a_{n+1-k}}{k}\frac{-1}{k+1}=\frac{1}{(n+1)(n+2)}+\sum_{k=1}^{n}\frac{a_{n+1-k}}{k}\frac{-1}{k+1}$
Si $1\leq k\leq n$ tenemos que $n+1\geq k+1\implies\frac{1}{n+1}\leq\frac{1}{k+1}\implies\frac{-1}{n+1}\geq\frac{-1}{k+1}\implies\frac{a_{n+1-k}}{k}\frac{-1}{k+1}\leq\frac{a_{n+1-k}}{k}\frac{-1}{n+1}$ (ya que $1\leq n+1-k\leq n\implies a_{n+1-k}>0$, y $k>0$), y como $\sum_{k=1}^{n+1}\frac{a_{n+1-k}}{k}=0=\frac{-1}{n+1}+\sum_{k=1}^{n} \frac{a_{n+1-k}}{k}\implies\sum_{k=1}^{n}\frac{a_{n+1-k}}{k}=\frac{1}{n+1}$, obtenemos que $\sum_{k=1}^{n+1} \frac{a_{n+1-k}}{k+1}=\frac{1}{(n+1)(n+2)}+\sum_{k=1}^{n}\frac{a_{n+1-k}}{k}\frac{-1}{k+1}\leq\frac{1}{(n+1)(n+2)}+\sum_{k=1}^{n}\frac{a_{n+1-k}}{k}\frac{-1}{n+1}=\frac{1}{(n+1)(n+2)}+\frac{-1}{(n+1)^2}=\frac{-1}{(n+1)^2(n+2)}<0$, que era lo que queríamos demostrar.