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.


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:

enter image description here


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.

  • 1
    $\begingroup$ 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? $\endgroup$ – user119615 Jan 5 '15 at 19:14
  • $\begingroup$ 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). $\endgroup$ – BaronVT Jan 5 '15 at 19:22
  • 1
    $\begingroup$ Ok, thanks, it was still a cool argument. $\endgroup$ – user119615 Jan 5 '15 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.