Problem 2023

Solution:

For $n = 1$ we have $a_{1} + \frac{1}{2} a_{0} = 0$, so $a_{1} = \frac{1}{2} > 0$.

Now suppose $a_{i} > 0$ for $1 \leq i < n$. From what is given it follows that
$$\sum_{k=1}^{n-1} \frac{a_{n-k}}{k+1} + \frac{a_{n}}{1} + \frac{a_{0}}{n+1} = 0 \Leftrightarrow a_{n} = \frac{1}{n+1} - \sum_{k=1}^{n-1} \frac{a_{n-k}}{k+1} \quad (*)$$
and likewise
$$0 = \sum_{k=0}^{n-1} \frac{a_{(n-1)-k}}{k+1} = \sum_{k=1}^{n} \frac{a_{n-k}}{k} = \sum_{k=1}^{n-1} \frac{a_{n-k}}{k} + \frac{a_{0}}{n} \Leftrightarrow \frac{1}{n} = \sum_{k=1}^{n-1} \frac{a_{n-k}}{k}.$$
This leads to
$$\frac{1}{n+1} = \sum_{k=1}^{n-1} \left( \frac{a_{n-k}}{k} \cdot \frac{n}{n+1} \right)$$
and, because of $n > k \Leftrightarrow \frac{n}{n+1} > \frac{k}{k+1} \Leftrightarrow \frac{n}{k(n+1)} > \frac{1}{k+1}$, to
$$\frac{1}{n+1} > \sum_{k=1}^{n-1} \frac{a_{n-k}}{k+1}.$$
With $(*)$ this gives
$$a_{n} > \sum_{k=1}^{n-1} \frac{a_{n-k}}{k+1} - \sum_{k=1}^{n-1} \frac{a_{n-k}}{k+1} = 0.$$
By complete induction it therefore holds that $a_{n} > 0$ for all $n \geq 1$.