convergence of series implies convergence of coefficients Is it true that
$$\sum_{i=0}^\infty a_{i_n} y^i \rightarrow \sum_{i=0}^\infty a_{i} y^i \quad \forall y \in [0,1]$$
implies
$$a_{i_n} \to_{n \to \infty} a_{i} \quad \forall i$$
where $0 \leq a_i,a_{i_n} \leq 1$ for all $i,n \in \mathbb{N}$ and $\sum_{i=0}^\infty a_i = \sum_{i=0}^\infty a_{i_n} = 1 \quad \forall n \in \mathbb{N}$?

Choosing $y = 0$ immediately gives $a_{0_n} \to_{n \to \infty} a_{0}$

My idea (please help for correctness and rigorousness):
$$\operatorname{lim}_{n\to\infty} \sum_{i=0}^\infty a_{i_n} y^i = \sum_{i=0}^\infty a_{i} y^i \quad \forall y \in [0,1]$$
Since the series of coefficients converges and dominates this power series for all $y\in [0,1]$ we have
$$\implies \sum_{i=0}^\infty \operatorname{lim}_{n\to\infty} a_{i_n} y^i = \sum_{i=0}^\infty a_{i} y^i \quad \forall y \in [0,1]$$
Two power series are equal iff all coefficients are equal, thus we have:
$$\implies \operatorname{lim}_{n\to\infty} a_{i_n} = a_i$$
as desired.
Is the above argument valid?
 A: One possible way to show this is to consider orthogonal polynomials on $[0,1].$ Let $f_{n}=\sum_{i=0}^{\infty}a_{i_{n}}y^{i},$ for all $n\geq1$ and let $f=\sum_{i=0}^{\infty}a_{i}y^{i}.$ If we let $\{p_{n}\}_{n=0}^{\infty}$ be a collection of polynomials with $\mathrm{deg}(p_{n})=n$ for all $n\geq0,$ with the property that $$\langle p_{n},p_{m}\rangle=\int_{0}^{1}p_{n}(x)p_{m}(x)\mathrm{d}x=\begin{cases}1&\text{if }n=m\\0&\text{otherwise,}\end{cases}$$ then we have that if $g(y)=\sum_{i=0}^{\infty}b_{i}p_{i},$ we can cover the coefficients by $\langle g,p_{i}\rangle=b_{i},$ for all $i\geq0,$ and these maps are continuous. Since $f_{n}\rightarrow f$ pointwise, and since the $f_{n}$ and $f$ are continuous and bounded (so that we don't have to worry about these integrals being finite), we have $\langle f_{n},p_{i}\rangle\rightarrow\langle f,p_{i}\rangle$ for all $i\geq0.$
Now the important thing is that if $g(y)=\sum_{i=0}^{\infty}a_{i}y^{i}=\sum_{i=0}^{\infty}b_{i}p_{i},$ then $b_{i}=\sum_{j=0}^{i}c_{i,j}a_{j}$ for some constants $c_{i,j}$ which do not depend on $g,$ and this allows us to recover the $a_{i}$ coefficients from the $b_{i}$ coefficients. In short, in your problem above, the convergence of the $b_{i}$ (the $\langle f_{n},p_{i}\rangle$) implies the convergence of the $a_{i}$ (your $a_{i_{n}}$), which you could show by a strong induction on $i$, if you wanted to show all of the details ($b_{0}=c_{0,0}a_{0}$ where $c_{0,0}\neq0,$ so convergence of $b_{0}$ and $a_{0}$ are equivalent, and then $b_{1}=c_{1,0}a_{0}+c_{1,1}a_{1},$ where $c_{1,1}\neq0,$ so since $a_{0}$ and $b_{1}$ converge, $a_{1}$ converges, and so on).
