(12/15/2009, 01:17 AM)bo198214 Wrote: 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.
So then how come the series given in the original post
\( \sum_{n=0}^{x-1} e^{e^n} = x + \sum_{n=1}^{\infty} \frac{e^{nx} - 1}{n! \left(e^n - 1\right)} = -\left(\sum_{n=1}^{\infty} \frac{1}{n! \left(e^n - 1\right)}\right) + x + \left(\sum_{n=1}^{\infty} \frac{1}{n! \left(e^n - 1\right)} e^{nx}\right) \)
looks to converge? In fact, the two series on the right can be proven to converge by applying the direct comparison test between these and the original series without its initial term, at a given x-value for the second series (the one that depends on x) and at \( x = 0 \) for the first (the one that sums to a constant independent of x). The original is known to converge because the radius of convergence of the exponential function's Taylor series is infinity and it is a simple substitution of a finite value (for finite x) into that series. So can you shoot down this proof? Unless you can claim
\( \sum_{n=0}^{x-1} e^{nx} = \frac{e^{nx} - 1}{e^n - 1} \)
is wrong or inconsistent with the Faulhaber formula. How would you do that?
