# Is there a simple proof that a power series can be differentiated term by term?

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?

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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.