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The following theorem is in Rudin's Principles of Mathematical Analysis, 8.2:

Suppose $c_n \geq 0$ and $S = \sum c_n$ converges. Put $f(x) = \sum_{n=0}^{\infty} c_nx^n$. Show that

$$ \lim_{x \rightarrow 1} f(x) = \sum_{n=0}^{\infty} c_n. $$

Rudin goes on to give a somewhat involved proof using Abel Summation. I'm wondering why this is necessary; why not just this argument?

"Proof:" Pick $N$ such that $\sum_{n > N} c_n < \epsilon$. Pick $r$ such that $1-r^N < \epsilon/S$. Then we have

$$ \sum_{n=0}^{\infty} c_n - \sum_{n=0}^{\infty} c_nr^n = \sum_{n=0}^N c_n(1-r^n) + \sum_{n>N} c_n(1-r^n) < 2 \epsilon. $$

Briefly, Rudin's method involves using Abel summation and then bounding the new sequence of partial sums, which seems like a less natural approach to me. Does anyone have a reason why he did that?

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up vote 4 down vote accepted

In my copy of Rudin there is no assumption that $c_n \ge 0$, and your proof fails if you can't assume that $\sum c_n$ converges absolutely.

Indeed, the power of this theorem is precisely that it applies in situations where $\sum c_n$ does not converge absolutely. For example it applies to $\arctan x = x - \frac{x}{3} + \frac{x}{5} \mp ... $ to give the classic formula $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} \mp ...$.

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I see. In my problem set, the assumption c_n > 0 is listed but now that I look at it, it's not in the book. So to make sure, trouble arises in my second term \sum_{n>N} c_n(1-r^n) actually being bounded by epsilon, right? – Tony Feb 24 '11 at 19:59
@Tony: yep. I think trying to fix the problem will probably eventually lead to some form of Abel summation anyway. – Qiaochu Yuan Feb 24 '11 at 20:06

I agree heartily with Qiaochu: the case in which all terms are non-negative is a trivial case of Abel's Theorem. This comes out in my take on Abel's Theorem in these notes on an undergraduate analysis class I taught a few years ago.

Note that I begin by giving exactly the proof from Rudin (why not?). After that I have some discussion of "Abel summation" and applications of this result. Note in particular Exercise 40: a series with non-negative terms is Abel-summable iff it is convergent in the usual sense.

It might also help to see some typical applications of Abel's Theorem, which are to non-convergent series. In my notes, I first use it to prove that if the Cauchy product of two convergent series converges at all, then it converges to the expected value, namely the product of the two sums. Note that earlier on in my notes I had proved that when at least one of the two series is absolutely convergent the Cauchy product necessarily converges to the expected value, whereas there are examples of divergent Cauchy products of two nonabsolutely convergent series. Then I use Abel's Theorem to justify the "calculus identity"

$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}$.

In both cases the setting is of non-absolute convergence.

I haven't looked back at my copy of Rudin, but I am willing to believe that he does not explicitly mention the significance of the non-absolute convergence here. If so, this is an instance of his book being a little too laconic for contemporary eyes.

Added: Okay, I just looked back at Rudin's book. First, he also follows the statement and proof of the theorem with the application to Cauchy products (what a coincidence...). Second -- it's true: he really should hit a little harder the point that the content comes when the terms are not all non-negative, and moreover when the series is not absolutely convergent. (Nevertheless Chapter 8 is by far my favorite in Rudin's book: a lot of classical content is covered in a very clean, efficient way.)

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What Qiaochu mentions is certainly correct, and probably the real reason. But here is another possibility (what I originally answered):

Suppose that $f(x)$ is a complex function on the unit disk. Your proof, while nice and quick for the real line, does not seem to work for any limit inside the disk. That is, sequences of the form $z_i$ with $|z_i|<1$ and $z_i\rightarrow 1$. However, the proof in Rudin does work in this case, and does not rely on the assumption that $x$ is real. (The last two lines do, but they can be modified to the complex case by simply putting in absolute values.)

Hope that helps,

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I don't think Rudin assumes x complex in this chapter. – Qiaochu Yuan Feb 24 '11 at 19:49
@Qiaochu: He doesn't, but sometimes authors put proofs that work in more general settings, so I thought that was the case. I didn't see the condition $c_n\geq 0$ was removed.... which certainly is a better reason – Eric Naslund Feb 24 '11 at 19:53
Even in the complex case, if the $c_n$'s were positive then Rudin's proof would be overkill; the series would converge absolutely hence uniformly on the closed disk, so it would be continuous there. – Jonas Meyer Feb 25 '11 at 5:08

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