Studying Analysis with 'Principles of Mathematical Analysis Third edition' written by Walter Rudin, I got some trouble from the proof of Taylor's theorem.

The theorem and the proof from the book are these: Theorem: Suppose $$f(x) = \sum_{n=0}^\infty c_n x^n,$$ the series converging in $|x|<R$. If $-R<a<R$, then $f$ can be expanded in a power series about the point $x=a$ which converges in $|x-a|<R-|a|$, and $$f(x)= \sum_{n=0}^\infty \frac{f^n(a)}{n!}(x-a)^n \quad (|x-a|<R-|a|).$$

Proof. We have $\begin{align*} f(x)&=\sum_{n=0}^\infty c_n[(x-a)+a]^n\\ &=\sum_{n=0}^\infty c_n \sum_{m=0}^n \binom{n}{m} a^{n-m}(x-a)^m\\ &=\sum_{m=0}^\infty \sum_{n=m}^\infty \binom{n}{m} c_n a^{n-m}(x-a)^{m}, \end{align*}$ and the equation is permissible if $$\sum_{n=0}^\infty \sum_{m=0}^n |c_n\binom{n}{m} a^{n-m}(x-a)^m|$$ converges, which is the same as $$\sum_{n=0}^\infty |c_n|(|x-a|+|a|)^n,$$ converging if $|x-a|+|a|<R.$

What I wonder is, if $\{c_n\}$ converges but not absolutely, for example $$c_n = (-1)^n\frac{1}{n},$$ isn't it troublesome convincing $$\sum_{n=0}^\infty |c_n|(|x-a|+|a|)^n$$ converges if $|x-a|+|a|<R$?


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