Prove or disprove: A power series that converges on $\left[0,1\right]$ is uniquely determined by any sequence of points converging to $1$.
More precisely, let $f, g:\left[0,1\right]\rightarrow\left[0,\infty\right)$ by expandable into power series $\begin{align}\forall x\in\left[0,c\right] &\bullet f\left(x\right)=\sum_{i=0}^\infty a_ix^i\\ \forall x\in\left[0,c\right] &\bullet g\left(x\right)=\sum_{i=0}^\infty b_i x^j\end{align}$ with non-negative coefficients that sum to $1$: $\left\{\left.a_i, b_i\space:\right|i\in\mathbb{N}_0\right\}\subseteq\left[0,\infty\right)$ $\sum_{i=0}^\infty a_i = 1 = \sum_{i=0}^\infty b_i$
Let $\left(s_j\right)_{j=0}^\infty$ be a strictly ascending sequence of points in $\left(0,1\right)$ that converges to $1$.
Is it true that $\left(\forall j\in\mathbb{N}_0\bullet f\left(s_j\right)=g\left(s_j\right)\right)\implies\left(\forall i\in\mathbb{N}_0\bullet a_i=b_i\right)$
Notes
If the radii of convergence $R_f, R_g$ were both $>1$, or if $\left( s_j\right)_{j=0}^\infty$ converged to any number $\in\left[0,1\right)$, the claim would be an immediate consequence of Baby Rudin's Theorem 8.5:
Suppose the series $\sum a_n x^n$ and $\sum b_n x^n$ converge in the segment $S=\left(-R, R\right)$. Let $E$ be the set of all $x\in S$ at which $\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty b_n x^n$ If $E$ has a limit point in $S$, then $a_n=b_n$ for $n=0,1,2,\dots$
The claim derives from Klenke, where it is stated as true without proof. (Theorem 3.2(iii))