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Why does the sum rule of differentiation fail sometimes for an infinite sum.(In terms of the derivation of the sum rule.)An example where it fails is here: Interchanging the order of differentiation and summation .

Vivaan Daga
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Infinity breaks everything and it takes some work to fix stuff. For instance, even simple things like associativity and commutativity are broken. What is

$$\sum_{n=1}^{\infty} (-1)^n ?$$

Associate one way and you get:

$$(-1+1) + (-1+1) + \cdots = 0 + 0 + \cdots =0.$$

Associate another way and you get

$$ -1 +(1-1) + (1-1) + \cdots = -1 +0+0\cdots =-1.$$

And $0 \neq -1,$ so we have a problem. Commutativity also breaks. Since you have infinite $1$'s and $-1$'s you can rearrange and get any number you like. Say $5$:

$$1+1+1+1+1+(-1+1)+(-1+1)+\cdots = 5.$$

So the point is that all the rules of math that work for finite things may not work for infinite things.

How do we fix this? Well, we have "absolute convergence" and "uniform convergence." If your series is absolutely convergent, then associativity, commutativity are restored. If your series (of functions) is uniformly convergent your sum rule for differentiation are restored.

So the answer to your question is "Because your series isn't uniformly convergent, so all the math that we used to derive the sum rule doesn't work anymore."

B. Goddard
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  • Does uniform convergence imply absolute convergence? – Vivaan Daga Jun 02 '20 at 13:06
  • Off the top of my head, I don't know. Absolute convergence is at a single point. Uniform convergence is for sequences of functions and happens on an interval (or domain.) So they're sort of different issues. I would bet there is an example of a series of functions that converges uniformly on an interval, but at one of the endpoints, the convergence is conditional. I can't think of one though. – B. Goddard Jun 02 '20 at 13:18
  • Here we go: https://math.stackexchange.com/q/126015/362009 – B. Goddard Jun 02 '20 at 13:20
  • So associativity , commutativity may fail for a series that is uniformly convergent – Vivaan Daga Jun 02 '20 at 13:40
  • My guess is yes. If a series converges conditionally, as in the example in that link, then you can re-arrange it to add up to just about anything. For the derivative, you need the arithmetic in the definition of derivative to hold. – B. Goddard Jun 02 '20 at 14:03
  • Don’t you need uniform convergence for the series of the sequence of derivatives? – Vivaan Daga Jun 27 '20 at 18:41
  • @VivaanDaga I think I said that. – B. Goddard Jun 27 '20 at 18:44
  • The series of derivatives I think has to be uniformly convergent not the series or am I wrong if so what is the exact theorem? – Vivaan Daga Jun 27 '20 at 18:50