Show that $g(x)$ is identically $0$ on an interval of convergence

Following Did's hint . . .

Suppose $g(x)$ is not identically $0$ on $(-R,R)$.

Then the coefficients$\;b_0,b_1,b_2,...\;$ are not all zero.

Let $k$ be the least nonnegative integer such that $b_k \ne 0$.

Then on $(-R,R)$, $$g(x) = x^kf(x)$$ where $$f(x) = \sum_{n=k}^{\infty}b_nx^{n-k}$$

Since $g$ converges on $(-R,R)$, so does $f$, hence $f$ is continuous on $(-R,R)$.

For all $n$, \begin{align*} &g(x_n) = 0\\[4pt] \implies\;&(x_n)^kf(x_n) = 0\\[4pt] \implies\;&f(x_n) = 0&&\text{[since $x_n \ne 0$]}\\[4pt] \end{align*} But then since $f$ is continuous on $(-R,R)$, $$b_k = f(0) = \lim_{n \to \infty}f(x_n) = \lim_{n \to \infty} 0 = 0$$ contradiction.