Assume I have a sequence $f_n$ of power series of the form $$ f_n(x) = \sum_{i=0}^\infty{a_{n,i}x^i},\quad a_{n,i}=\begin{cases}\alpha_{n,i} & n\leq i,\\b_i & n>i.\end{cases}.\tag{*} $$
This implies, in particular, that $a_{n,i}\xrightarrow{n\to\infty}b_i$, for each $i$, even though this convergence might not be uniform.
Let us further assume that the power series $f_n$ converge uniformly on $[0,1]$ to a limit $f$. It is my understanding that simple convergence of coefficients is not enough to conclude convergence of a series of power series to the power series corresponding to the limiting coefficients. It does seem, however, that if the power series $g(x)=\sum_i{b_i x^i}$ has positive radius of convergence $r$, then at least for $0<x<r$, the functions $f=\lim f_n$ and $g$ coincide.
Question: Can the limit $f=\lim_{n\to\infty} f_n$ of power series of the form (*) be described in terms of the limiting coefficients $b_i$. If yes, how?
This is false. To see this, recall the Stone-Weierstrass theorem (http://en.m.wikipedia.org/wiki/Stone–Weierstrass_theorem#Locally_compact_version).
We will apply this to the locally compact space $X=(0,1]$. It is easy to see that $x \mapsto x \in C_0(X)$. Now apply Stone-Weierstrass in the version above with the algebra generated by $x^n$ (i.e. the set $A_n = \{\sum_{k=1}^N x^{kn}\mid N \in \Bbb{N}, a_1,\dots a_N \in \Bbb{R}\}$) for $n\geq 10$ fixed. Observe that $A_n$ indeed separates the points of $X$ and that the sum begins at $k=1$, not $k=0$.
This implies that there is a polynomial $p_n \in A_n$ such that $\sup_{x\in X} |x - p_n (x)| < 1/n$.
Now set $f_n (x) := x -p_n(x)$ and observe that continuity of $f_n$ on $I= [0,1]$ together with the convergence above and with density of $X$ in $I$ implies $f_n \to 0 =: f$ uniformly.
But the sequence $f_n$ satisfies your requirements with $b_1 = 1$ and $b_k =0$ for $k\neq 1$, i.e. for $g( x)=x$. But $g\neq f$.