Proof that real power series is real analytic

I'm wondering if the following argument is correct. The proof in the book is longer and I don't understand it.

Theorem. Suppose $f(x) = \sum_{n=0}^\infty a_n x^n$, where the series converges for $-R < x < R$. Then for each $c \in (-R,R)$, there is a sequence $(b_k)$ such that $f(x) = \sum_{k=0}^\infty b_k (x-c)^k$, converging for $\left|x -c\right| < R - \left|c\right|$.

Proof. Let $c \in (-R,R)$. If $c = 0$ the result is trivial. Suppose that $c \ne 0$ and that $\left|x -c\right| < R - \left|c\right|$. Then $\left|x\right| \le \left|x-c\right| + \left|c\right| < R$, so $\sum_{n=0}^\infty a_n x^n$ converges absolutely. Notice that $$x^n = (x-c + c)^n = \sum_{k=0}^n\binom{n}{k}c^{n-k}(x-c)^k = \sum_{k=0}^\infty \binom{n}{k}c^{n-k}(x-c)^k.$$ Therefore $$\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty a_n \sum_{k=0}^\infty \binom{n}{k}c^{n-k}(x-c)^k = \sum_{k=0}^\infty \sum_{n=0}^\infty a_n \binom{n}{k}c^{n-k}(x-c)^k,$$ where the order of the summation was interchanged by the absolute convergence of $\sum_{n=0}^\infty a_n x^n$. This proves the theorem with $b_k = \sum_{n=0}^\infty a_n \binom{n}{k}c^{n-k}$. $\square$

• You are a bit short with the argument by absolute convergence, but your argument is valid - assuming you have already proven that the convergence of a power series is absolute in the interior of the interval/disk of convergence. (You probably have, going by the way you cite the fact.) – Daniel Fischer May 28 '15 at 17:28
• $|x-c|< R-|c|< R$, so $\sum a_n (x-c)^n$ converges. So there is of course a sequence $(b_k)$, viz. $(a_k)$, such that $\sum b_k (x-c)^k$ converges! – KalEl May 28 '15 at 17:35
• @DanielFischer Yes, it's an easy consequence of the root test. Thanks for your comment. – Randy Randerson May 28 '15 at 17:58
• @KalEl The series must also be equal to $f(x)$, so I don't think your argument proves the result. – Randy Randerson May 28 '15 at 17:59