# 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: ## 3 Answers

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 $0\leq x<1$.

If $\displaystyle \sum_{n=0}^\infty a_n$ converges then $a_n \to 0$ and $|a_n| \to 0$,

so $|a_n x^n|$ is bounded above by $\max_k(|a_k|) |x|^n$,

and thus the partial sum $\displaystyle \sum_{n=0}^m |a_n x^n|$ is an increasing sequence bounded above by $\dfrac{\max_k(|a_k|)}{1-|x|}$ if $0 \le x \lt 1$,

making the infinite series absolutely convergent.

You can easily extend this to $-1 \lt x \lt 1$.

If the limit $\lim_{n\to\infty}\left| \dfrac{a_{n+1}}{a_n} \right|$ exists, then by the ratio test, if

$$\lim_{n\to\infty}\left| \dfrac{a_{n+1}}{a_n} \right| > 1$$

then the original series diverges, which we know to be false. Thus the ratio above is either $1$ or less than $1$.

Therefore, when applying the ratio test to the power series, we examine

$$\lim_{n\to\infty}\left| \dfrac{a_{n+1}x^{n+1}}{a_nx^n} \right| = \left| \dfrac{a_{n+1}x}{a_n} \right| < \left| \dfrac{a_{n+1}}{a_n} \right| \leq 1$$

for $|x| < 1$, and we have absolute convergence.

If the limit doesn't exist, then the statement of the ratio test involves the lim inf and lim sup, and this argument doesn't work; thus the other answers given ($|a_n x^n| \leq \max_k |a_k| x^n$) are more complete.

• I just have a follow-up question to this. Even if the original series converges, can we still use the argument with the ratio-test? I mean, the ratio-test says that if they ratio is so and so, then the series converges. But you use the converse, I mean, could there be a case where a series converges, but the ratio-limit might fail to exist? – user119615 Jan 5 '15 at 19:14
• Yes, this is possible; in this case, if the lim inf of the ratio is $>1$, you get divergence, and convergence if the lim sup $< 1$. In view of this, I don't think my answer works if the limit doesn't exist, and so the other answers are better in this regard (see my edits above reflecting this). – BaronVT Jan 5 '15 at 19:22
• Ok, thanks, it was still a cool argument. – user119615 Jan 5 '15 at 19:32