I am trying to solve this [Komal problem 661](https://www.komal.hu/feladat?a=feladat&f=A661&l=en):



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> 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 $n\ge 1$.
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> 
> 


**Add edit**:For the Iosif Pinelis point out,$b\_{0}=(-1)^K$,Now I have known where my wrong,and Ira Gessel point that, Now How to prove


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)^K\\
\implies \left(\sum\_{i\geq 0} a\_i x^i\right)^K&=\frac{(-1)^K}{g(x)}=\frac{-(-1)^K\cdot x}{\sum\_{i\geq 1} -\cfrac{x^{i}}{i}}=\frac{-(-1)^K\cdot x}{\ln(1-x)}\\
\implies\sum\_{i\geq 1} a\_i x^i &=-a\_{0}+ \left(\frac{-(-1)^Kx}{\ln(1-x)}\right)^{1/K}
=1-\left(\dfrac{-x}{\ln{(1-x)}}\right)^{1/K}\end{align}
Thus, using the Tyalor series expansion,


$$\dfrac{d^i}{dx^i}\left(1-\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}\right)|\_{x=0}>0\tag{1}$$


But the last maybe it not easy prove it,can help me to prove $(1)$?Thanks



show this inequality with $\frac{d^i}{dx^i}\left(1-\left(\frac{-x}{\ln(1-x)}\right)^{1/K}\right) \Bigg|_{x=0}>0, ~~~\forall i\in N^{+}$
The function $f(x) = (x - 1) / \log x$ extends to a holomorphic function on $\mathbb C \setminus (-\infty, 0]$. Clearly, $f(x) > 0$ when $x > 0$. We claim that $\operatorname{Im} f(z) \geqslant 0$ when $\operatorname{Im} z > 0$, that is, $f$ is a *complete Bernstein function*. For the proof of this claim, see the last two paragraphs of this answer.


Similarly, $g(x) = x^{1/K}$ is a complete Bernstein function: $g(x) > 0$ for $x > 0$, and $g$ extends to a holomorphic function on $\mathbb C \setminus (-\infty, 0]$, and we have $\operatorname{Im} g(z) \geqslant 0$ when $\operatorname{Im} z > 0$.


It follows that $h(x) = g(f(x))$ is a complete Bernstein function. In particular, $h$ is a Bernstein function: $(-1)^{n - 1} h^{(n)}(x) \geqslant 0$ for $x > 0$ and $n \geqslant 1$. We conclude that $F(x) = -h(1 - x) = -(-x / \log(1 - x))^{1/K}$ has all derivatives nonnegative for $x < 1$: $F^{(n)}(x) \geqslant 0$ when $x < 1$ and $n \geqslant 0$, as desired.




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*Proof of the claim:* My favourite way to prove this kind of results is to observe that $\operatorname{Arg} \log z$ is a bounded harmonic function in the upper complex half-plane with boundary values $0$ on $(1, \infty)$, $\pi$ on $(0, 1)$ and something in $[0, \pi]$ on $(-\infty, 0)$. Similarly, $\operatorname{Arg} (z - 1)$ is a bounded harmonic function in the upper complex half-plane with boundary values $0$ on $(1, \infty)$ and $\pi$ on $(-\infty, 1)$. Thus, $$\operatorname{Arg} (z - 1) - \operatorname{Arg} \log z = \operatorname{Arg} f(z)$$ is a bounded harmonic function in the upper half-plane with boundary values $0$ on $(0, \infty)$ and something in $[0, \pi]$ on $(-\infty, 0)$. This proves that $\operatorname{Arg} f(z)$ is well-defined and belongs to $[0, \pi]$ in the upper complex half-plane.


An alternative approach would be to write the representation of the logarithm as a *extended complete Bernstein function* (or simply a *Nevanlinna–Pick function*, to give another keyword):
$$ \log z = \int\_0^\infty \biggl( \frac{z}{z + s} - \frac{1}{1 + s} \biggr) \frac{1}{s} \, ds . $$
It follows that
$$ \tilde f(z) = \frac{z \log z}{z - 1} = \int\_0^\infty \frac{z}{z + s} \, \frac{s}{1 + s} \, \frac{1}{s} \, ds $$
is a complete Bernstein function, and consequently $f(z) = z / \tilde f(z)$ is a complete Bernstein function, too.


For more on Bernstein functions, complete Bernstein functions and related topics, see the excellent book *Bernstein Functions. Theory and Applications* by René Schilling, Renming Song and Zoran Vondraček.


