What allows us to differentiate closed form summation?

Asked Apr 02 '20 at 00:21

Active Apr 02 '20 at 00:21

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I've written a proof with some help from outside resources about expected number of coin tosses before observing a single head. The proof involves one part where I take the derivative of both the terms of summation and the closed form of an infinite geometric series. I understand the proof mostly, but why are we able to say "and now we just take the derivative of the summation..."

In other words, I'm asking, given that we know the summation of an infinite geometric series converges to its closed form, why can we "take the derivate" of whatever the summation is summing along with the derivative of the closed form and say they are still equal expressions? I hope that makes sense.

asked Apr 02 '20 at 00:21

dover

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The propert you're looking for is uniform convergence. – CoffeeCrow Apr 02 '20 at 00:24
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You can prove the radius of convergence of the derivative is the same as the radius of convergence of the series as is done in this post. https://math.stackexchange.com/questions/1914354/showing-the-radius-of-convergence-for-a-power-series-is-equal-to-the-radius-of-c?rq=1 – CyclotomicField Apr 02 '20 at 00:30
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@dover You may find reading Chapter 8 Uniform Convergence and differentiation, especially section $8.5$, "Differentiating A Sequence", interesting and possibly helpful. – John Omielan Apr 02 '20 at 00:40
Hey gang thanks for the recommendations! Yeah this one has been bothering me for some time, I expected it would require reading. – dover Apr 02 '20 at 00:45

What allows us to differentiate closed form summation?

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