Let $K$ be a fixed positive integer,Let $(a_{0},a_{1},\cdots )$ be the sequence of real numbers that satisfies $a_{0}=-1$ and $$\sum_{i_{0},i_{1},\cdots,i_{K}\ge 0,i_{0}+i_{1}+\cdots+i_{K}=n}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}=0$$ for every postive integer $n$,show that $a_{n}>0$ for $n\ge 1$
This problem is from Komal problem 661. https://www.komal.hu/feladat?a=feladat&f=A661&l=en
Continuation of the idea suggested by Vlad Matei
Let $\displaystyle f(x)\triangleq\sum_{i\geq 0} a_i x^i$ and $\displaystyle g(x)\triangleq \sum_{i\geq 0} \cfrac{x^{i}}{i+1}$.
Then, we get $$ f(x)^Kg(x) = \sum_{n\geq 0}b_nx^n \text{ with } b_n=\sum_{\substack{i_{0},i_{1},\cdots,i_{K}\ge 0\\i_{0}+i_{1}+\cdots+i_{K}=n}}\dfrac{a_{i_{1}}a_{i_{2}}\cdots a_{i_{K}}}{i_{0}+1}.$$ Since $b_n=0$ for $n\geq 1$, we get \begin{align} f(x)^Kg(x)&=b_0=-1\\ \implies \left(\sum_{i\geq 0} a_i x^i\right)^K&=\frac{-1}{g(x)}=\frac{x}{\sum_{i\geq 1} -\cfrac{x^{i}}{i}}=\frac{x}{\ln(1-x)}\\ \implies\sum_{i\geq 0} a_i x^i &= \left(\frac{x}{\ln(1-x)}\right)^{1/K}. \end{align} Thus, using the Tyalor series expansion, $$ a_i=\frac{1}{i!}\frac{d^i}{dx^i}\left(\frac{x}{\ln(1-x)}\right)^{1/K} \Bigg|_{x=0}>0\iff \frac{d^i}{dx^i}\left(\frac{x}{\ln(1-x)}\right)^{1/K} \Bigg|_{x=0}>0.$$
This only a partial answer but here is a way to get an expression for the sequence. Define the generating function of the sequence $\displaystyle f(X)=\sum_{i\geq 0} a_i X^i$ and let $\displaystyle g(X)=\sum_{i\geq 0} \cfrac{X^{i}}{i+1}$. Then we can rewrite the recurrence as $(f(X))^Kg(X)=-1$. Thus $f(X)=-(g(X))^{-1/K}=-(1+h(X))^{-1/K}$. We now use the binomial theorem to get $\displaystyle f(X)=-1-\sum_{n\geq 1} \dbinom{-1/K}{n}(h(X))^n$. Equating the coefficients of $X^i$ on both sides we get $$a_i=-\sum_{\substack{s\geq 1,m_1,\ldots, m_s\geq 1 \\ m_1+\ldots+m_s=i}} \dbinom{-1/K}{i}\cdot \cfrac{1}{(m_1+1)\ldots(m_s+1)}$$ for $i\geq 1$. After this you need to do some bounding to show that the first term $\cfrac{1}{Ki}$ dominates the negative terms.
