Uniqueness of power series determined by a sequence of points converging to a boundary point 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))
 A: It's not true.  Consider the following series:
$$ \eqalign{f(x) &= \sum_{j=1}^\infty \frac{3}{2^{2j}} x^{n_{2j-1}}\cr
            g(x) &= \frac{1}{2} + \sum_{j=1}^\infty \frac{3}{2^{2j+1}} x^{n_{2j}}\cr}$$
where $n_j$ is an increasing sequence of positive integers to be chosen.
Thus 
$$h(x) = g(x) - f(x) = \frac{1}{2} + \frac{3}{2}\sum_{k=1}^\infty (-1)^k  \frac{x^{n_k}}{2^k}$$
Let $$h_N(x) = \frac{1}{2} + \frac{3}{2}\sum_{k=1}^N (-1)^k  \frac{x^{n_k}}{2^k}$$
Note that $(-1)^N h_N(1) = (1/2)^{N+1} > 0$.
I will choose the $n_j$ and an increasing sequence $t_j \in [0,1)$ inductively so that $(-1)^j h(t_j) > 0$.  This can be done as follows.  We start with $t_0 = 0$, so $h(t_0) = 1/2 > 0$.
Given $N$, $t_{N-1}$ and $n_1, \ldots, n_N$, take $t_N \in (t_{N-1},1)$ so close to $1$ that $(-1)^N h_N(t_N) > 0$.   Since $|h(t) - h_N(t)| < 3 t^{n_{N+1}}/2^{N+1}$, by taking
$n_{N+1}$ large enough we can ensure that $(-1)^N h(t_N) > 0$.
Now note that by the Intermediate Value Theorem, for each $N$ there is $s_N$ with $t_N < s_N < t_{N+1}$ such that $h(s_N) = 0$.
