Given nonnegative sequences $(b_j)_{j\ge0}$ and $(u_j)_{j\ge0}$ with $u_0=b_0=1$ and $\sum_{j=0}^{n}u_{j}b_{n-j}=1$ for each $n\ge1$. Let $$ b(x)=\sum_{j=1}^{\infty}(b_{j-1}-b_{j})x^{j}. $$ Suppose that $\frac{1}{1-b(x)}=\sum_{j=0}^{\infty}a_jx^{j}$. I want to find that coefficient of the expansion, i.e. $a_j$. I think $a_j=u_j$ for each $j$, and, by using the given condition $\sum_{j=0}^{n}u_{j}b_{n-j}=1$, I have found that $a_1=u_1$, $a_2=u_2$ and $a_3=u_3$. For example, \begin{align} \frac{1}{1-b(x)}&=1+b(x)+\left(b(x)\right)^2+\dots\\ &=1+\left((1-b_1)x+(b_1-b_2)x^2+\dots\right)+\left(b(x)\right)^2+\dots, \end{align} from which it follows that $a_1=1-b_1$. On the other hand, we have from the condition (when $n=1$) that $1=u_0b_1+u_1b_0=b_1+u_1$ implying $u_1=1-b_1$. Thus, $a_1=u_1$.
However, I couldn't come up with a better (a systematic) method to prove in general $a_j=u_j$ for all $j\ge1$. Is there any reference of this? Any help or hint would be greatly appreciated. Thank you very much.
Consider $$f(x):=\bigl(1-b(x)\bigr)\cdot \sum_{j=0}^\infty u_jx^j=\Bigl(1+\sum_{i=1}^\infty(b_i-b_{i-1}) x^i \Bigr) \cdot \sum_{j=0}^\infty u_jx^j \,.$$ The constant coefficient in the product $f(x)$ is $u_0 =1$. For $n \ge 1$, the coefficient of $x^n$ in $f(x)$ is $$ u_n+ \sum_{j=0}^{n-1} u_j (b_{n-j}-b_{n-j-1})= \sum_{j=0}^{n} u_j b_{n-j}-\sum_{j=0}^{n-1} u_j b_{n-j-1}=1-1=0 \,.$$ thus $f(x)\equiv1$ as needed.