Power Series Equality Proof

1. Jun 21, 2012

pendesu

1. The problem statement, all variables and given/known data
Given $\overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}$ and $\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n}$ that are in R. Then, $\overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}=\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n}$ if and only if $a_{n}=b_{n}$ for every $n=0,1,2,...$

The attempt at a solution
(<<) Assume an=bn for every n=0,1,2,...
Then $\overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}=a_{0}+a_{1}(x-a)+...=b_{0}+b_{1}(x-a)+...=\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n}$.
(>>) Now with this direction I am having some issues as the series may not necessarily converge. My attempts have been feeble at best. The problem I have had is that the sums are infinite so I don't think I can use that polynomials are equal if and only if their coefficients are equal. At this point I am thinking about looking at the nth partial sums but I am not sure and just would like any advice.

2. Jun 21, 2012

SammyS

Staff Emeritus
How can it be said that $\sum_{n=0}^\infty a_{n}(x-a)^{n}=\sum_{n=0}^\infty b_{n}(x-a)^{n}\ ,$ unless both sums converge?

3. Jun 21, 2012

pendesu

That is the thing. This was something I was going to ask my professor that I am going to be doing research with in p-adic analysis. I am starting to think this may just be a definition since if I recall my professor he was saying how a p-adic integer which is a power series may not be a convergent power series in the ring of p-adic integers. My professor is busy right now. This might not make sense.