Almost 3 months ago, I asked this question regarding if it's possible to compute the summation of derivatives, as in the example I've given: $$\sum_{n = 0}^\infty \frac{d}{dx} x^n$$ One answer regarded the interchange between summations and derivatives, which got me thinking: does the interchange between the derivative and the summation succeed in this example? In other words, is $$\frac{d}{dx}\left(\sum_{n = 0}^\infty x^n\right) = \sum_{n = 0}^\infty\left(\frac{d}{dx} x^n \right)$$ true? I believe it is, because the summation of the derivatives of $x^n$ from $n = 0 \to \infty$ was: $$1 + 2x + 3x^2 + 4x^3 + 5x^4 + \cdot \cdot \cdot$$ and to evaluate the summation of a series, you take the derivative of each term, which gets me: $$\frac{d}{dx}\left(\sum_{n=0}^\infty x^n\right) = \frac{d}{dx}(1 + x^2 +x^3 + x^4 + x^5 + \cdot \cdot \cdot) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \cdot \cdot \cdot $$ Hence, I believe that the interchange succeeds. Am I right? Does the interchange succeed?
Notes
- I implemented the left hand side of the "interchange equation" into WolframAlpha, and I got back something "useful", but it doesn't really solve my problem.
- I found This question and this question, but they have nothing to do with my question.
- Multiple other questions deal with interchanges with summations and integrals. This is about interchanging summations and derivatives.