real analysis - Why is the convergence absolute?

There is one thing my book uses in a proof after Abels theorem which I do not understand:

Lets say that $\sum_{n=0}^\infty a_n$ converges.

For $0\le x<1$, we look at $\sum_{n=0}^\infty a_n x^n$. The book says that this series converges absolutely for all the values of $x$ we have defined it. But why? We started with a sequence that might not even converge absolutely. And how do we know that we even have convergence when we introduce the $x$ variable? It would have been easy to see convergence if the orginal series was absolutely convergent, not only convergent, but they only state that the original series is convergent.

UPDATE:

If you are interested I got the picture from the book. Theorem 8.2 is Abels theorem, Definition 3.48 is the Cauchy-product(but this is clear from the picture), and what Theorem 3.51 states is also clear from the picture:

Answer

The convergence is absolute, provided that $0\leq x<1$. In other words, $x=1$ is not allowed (and Abel's theorem attempts to assign a sensible value to the sum at $x=1$). Since $\sum_{n=0}^\infty a_n$ converges, then $a_n\rightarrow 0$, which means that $|a_n|\leq C$ and so $\sum_{n\geq 0} |a_nx^n|\leq C \sum_{n\geq 0} |x|^n<\infty$ for $0\leq x<1$.