I know that the derivative of the sum of a power series can be calculated by summing the derivatives of the terms, and that the resulting series has the same radius of convergence as the original. The proofs I've seen, however, have either been quite long and involved, relying on several lemmas (e.g., the one on ProofWiki) or waved their hands in some difficult spots. Does anyone know of a simpler approach?

  • $\begingroup$ Isn't it easy with the $n$-th root test for the radius of convergence? That's the important thing to show, I believe (provided that you already know how differentiation and uniform convergence interact) $\endgroup$ Sep 4, 2013 at 19:56
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    $\begingroup$ Anything involving switching two limits (such as this proof) will either be somewhat technical or involve some handwaving, by its very nature. $\endgroup$
    – treble
    Sep 4, 2013 at 19:58
  • $\begingroup$ @GiuseppeNegro, no, I don't know how differentiation and uniform convergence interact, so I'll start looking in that direction. The $n$th root test is one of those lemmas... $\endgroup$
    – dfeuer
    Sep 4, 2013 at 20:11
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    $\begingroup$ It's pretty easy to show that the radius of convergence of the series obtained by termwise differentiation is at least as large as that of the original series. Then one can for example integrate, using the uniform convergence on the path of integration to interchange summation and integration. One can also prove it via direct manipulation of power series, but iirc, that is a little icky. Of course, one can also use the Cauchy integral, which gives a nice simple proof. But that uses some heavy machinery, of course. $\endgroup$ Sep 4, 2013 at 20:21
  • $\begingroup$ Before you go differentiating anything, you had better be sure the series converges uniformly in your domain. If it does not the derivative of the series does not need to be the series of the derivatives. To differentiate the entire series you have to take a difference. To get to the individual terms you have to switch the difference and the sum. Without uniform convergence they might not be equal. $\endgroup$
    – Betty Mock
    Sep 5, 2013 at 2:36

1 Answer 1


Note: I would post a summary of this as a comment, but I don't seem to have the rep.

Note 2: I'm not sure what your background is, so apologies if the beginning of this seems patronizing?

Anyway. Often in math, the "right" way to see a result is to understand the factors around it. It's all well and good to know that you can differentiate a power series term by term, but there are really only two ways to see why. Most obviously, you can prove it directly from the definitions, but if the proof is long, you may be convinced that the theorem is true, without having any sense of why it is true. The better way here (in my opinion) is to see why it is true, using two facts. You can see them as lemmas, but they're actually important in their own right.

First: within compact sets inside the radius of convergence, power series converge uniformly (you can see this from the root test and a geometric series argument). For now just think of compact subsets as closed intervals; that's all you need for this.

Second: if $\{f_n\}$ is a sequence of differentiable functions and converges uniformly to $f$ on an open set and $\{f_n'\}$ converges uniformly to $g$, then $f$ is differentiable on the open set and $f'=g$.

The first fact is definitely useful for a lot of purposes. Maybe that second fact is not so generally useful, though it gives you a sense of what's going on in this spot. But it's easy to prove the intended statement from these two facts:

Step 1: The partial sums of your power series are a sequence of differentiable functions which converge uniformly to $f$ (fact 1).

Step 2: Their derivatives form a power series which has the same radius of convergence. You should check this, but it isn't hard; it's a similar argument to the proof of fact 1. Therefore they converge uniformly to some function $g$.

Step 3: It must now be that $f'=g$ (fact 2). More simply, that the term-by-term derivative converges uniformly to $f'$ inside the radius of convergence. And you're done!

Hopefully this helps it seem like a more concise proof. You can't get away from fact 1 (you need to know it generally) and if you like, you can see fact 2 as the "technical part" of the proof, though it's a good exercise to get your hands dirty with uniform convergence.

  • $\begingroup$ I have some work ahead of me to get through all the pieces of this answer, but step 3 looks a bit shady at the beginning--it's not intuitively obvious that $f$ must be differentiable. $\endgroup$
    – dfeuer
    Sep 4, 2013 at 21:04
  • $\begingroup$ Just a minor note on step 1: perhaps it would be better to explicitly point out that the convergence is uniform on compact subsets within the radius of convergence. $\endgroup$ Sep 4, 2013 at 23:21
  • $\begingroup$ @GiuseppeNegro, that would be better, because I was wondering about that myself. It seems rather unlikely that the power series for the exponential function converges uniformly. $\endgroup$
    – dfeuer
    Sep 5, 2013 at 1:25
  • $\begingroup$ @dfeur: in my mind, we were starting with a function and finding its Taylor series. For an arbitrary power series there is another step which is perhaps nontrivial. And Giuseppe, you are correct. I will rephrase. $\endgroup$ Sep 5, 2013 at 1:49
  • $\begingroup$ Ah! I stated fact two too narrowly. So this avoids all the issues. Editing now. $\endgroup$ Sep 5, 2013 at 1:56

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