(12/14/2009, 09:22 PM)mike3 Wrote: How does that refute the summation I gave with the transseries? That involves continuum-summing on terms of the form \( ae^{nx} \), not \( ae^{nx^2} \) or \( ae^{x^n} \) or something like that.
Its because \( \frac{x^2}{2}<e^x \).
Thatswhy both Faulhaber sums are also in the < relation.
Thatswhy the sum of \( e^{\frac{x^2}{2}} \) and of \( e^{e^x} \) are also in this relation.
