How to prove Carrier's rule for divergent series?

1. Jul 12, 2008

Count Iblis

Carrier's Rule

2. Jul 12, 2008

HallsofIvy

Staff Emeritus
Did you see the word "humorous" in that description? There is no proof, its a joke.

3. Jul 12, 2008

Count Iblis

Yes, I know. But it is still the case that a few terms of an asymptotic series wil give you a good approximation while if you use a series that converges you typically need may terms to get a similar approximation.

This is then alleged to be caused by the fact that demanding convergence is an extra burden that comes at the expense of how good the approximation can be after only a few terms of the series.