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I can't figure out how it is valid to invoke the Absolute Convergence Theorem, whose hypothesis is "Let the power series have radius of convergence R", to establish case c of the Radius of Convergence Theorem (basically concluding that "the power series has a radius of convergence..."). Isn't it circular reasoning to prove X by assuming X itself?

(The aforementioned ACT is invoked beside the second, and subsequently again after the third, margin note below.) I would be extremely grateful for any help resolving this.

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P.S. The above are excerpted from David Brannan's Mathematical Analysis.

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It is a bit confusing the way he writes it, but what he actually shows in Theorem 3 (in the case $a=0$), is that if $\sum a_n x^n$ converges, then $\sum a_n y^n$ absolutely converges for any $|y|<|x|$. The invocation of $ACT$ is confusing since it speaks about a notion (radius of convergence) whose existence is proved in Theorem 1. However, in the proof of Theorem 3, $R$ is used only to take an $|x|<R$, so that we know $\sum a_n x^n$ converges. What he should have said is "from the proof of Theorem 3, etc...".

More details:

In the proof of Theorem 3 (in the book) he picks an arbitrary $|x|<R$ for which he has to show absolute convergence. Then he says there exists $X$ such that $|x|<|X|<R$. For this $X$ you have convergence, and from this convergence alone (no $R$ needed after this point) he deduces absolute convergence for $x$. Notice that at no point he assumes or uses that the series converges for $x$. He only uses radius of convergence to say that the series converges for $X$. In other words he proves the following: If $\sum a_n x^n$ is convergent, than for any $|y|<|x|$, $\sum a_n y^n$ is absolutely convergent.

For Theorem 1, IF a) and b) do not hold, there must be an $X$ for which the series diverges. He concludes that for any $|x|>|X|$, the series still diverges. Indeed, if it converges for some $|x|>|X|$, it would absolute converge for ANY $|y|<|x|$ (by the previous argument), in particular for $X$. Contradiction.

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@Ryan I edited the answer, perhaps now it is more clear and addresses your question. – Theo Mar 6 '13 at 16:56
Hey, thanks. I understand your explanation, but where did you get the proof of Theorem 3 from? I've just re-read it carefully, and what you say doesn't seem to be true; in particular, Thm 3 (and its proof) does seem to depend on the condition that a radius of convergence R exists, ie. that the series converges for all x such that |x|<R. If so, then my trouble with the argument of this current proof stands. – Ryan Mar 6 '13 at 17:57
@Ryan Please see further edits. – Theo Mar 7 '13 at 1:03
Hey. Just before the boldface, you wrote, "For this X you have convergence." This is exactly where I infer that the theorem depends on the existence of a radius of convergence (R). This is why we know that "For this X you have convergence", as the bit just quoted is neither an assumption nor a hypothesis/condition. – Ryan Mar 8 '13 at 9:59
@Ryan Yes, in the proof of Theorem 3 you have that from the existence of R. However, in the proof of Theorem 1, you have that the series converges for $x$ from the assumption by contradiction. IF (by contradiction)the series would converge for some $|x|>|X|$, it would also converge for all $|y|<|x|$, from the proof of T3. This $x$ is actually the capital $X$ in the proof of T3. – Theo Mar 12 '13 at 22:46

Here is an alternative which tells a little more, since it involves absolute convergence.

Assume that the power series converges at some $x_0\neq a$. That is: not a).

Then it converges absolutely for all $|x-a|<|x_0-a|$. Indeed, since $a_n(x_0-a)^n$ tends to $0$, $|a_n(x-a)^n|\leq C \left(\frac{|x-a|}{|x_0-a|}\right)^n$ for all $n$, so you can conclude by comparison with the geometric series on the right.

So we can consider the nonempty set $S'$ of all $R>0$ such that the series converges absolutely for all $|x-a|<R$.

If $R$ belongs to $S'$, then so does every $R'\leq R$ by comparison.

So we see that $S'$ is of one of these two forms:

  • case b): $(0,R]$ (you can check that the upper bound must belong to it). Note that for any $|x-a|>R$, the power series diverges, for otherwise $|x-a|$ would belong to $S'$ by the first remark above. So this is really case b).

  • case c): $(0,+\infty)$.

The radius of convergence is defined to be $R$ in the first case, $+\infty$ in the second case. And $0$ in case a).

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