I am trying to solve this Komal problem 661:

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 $n\ge 1$.

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

I am trying to solve this Komal problem 661:

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 $n\ge 1$.

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

I am trying to solve this [Komal problem 661]:

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 $n\ge 1$.

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, $$ a_i=\frac{1}{i!}\frac{d^i}{dx^i}\left(\frac{x}{\ln(1-x)}\right)^{1/K} \Bigg|_{x=0}>0\iff \frac{d^i}{dx^i}\left(\frac{x}{\ln(1-x)}\right)^{1/K} \Bigg|_{x=0}>0.$$

where this wrong?

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 $$\dfrac{d^i}{dx^i}\left(1-\left(-\dfrac{x}{\ln{(1-x)}}\right)^{\frac{1}{K}}\right)|_{x=0}>0$$ [Komal problem 661]:https://www.komal.hu/feladat?a=feladat&f=A661&l=en

I am trying to solve this Komal problem 661:

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 $n\ge 1$.

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