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Is it true that

$$\sum_{i=0}^\infty a_{i_n} y^i \rightarrow \sum_{i=0}^\infty a_{i} y^i \quad \forall y \in [0,1]$$


$$a_{i_n} \to_{n \to \infty} a_{i} \quad \forall i$$

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?

share|cite|improve this question
Stating it differently: We have $\sum_{i=0}^\infty (a_{i_n} - a_i)y^i \to 0 \quad \forall y \in [0,1]$. Can I somehow swap lim and sum with help of a congergent majorant and then conclude that the coefficients need to converge? – 193841 Oct 20 '13 at 17:07

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