03/24/2010, 07:43 PM
(03/24/2010, 05:57 PM)bo198214 Wrote: Hm, this works? This is really amazing, considering that
\( g_n=\frac{\gamma(A, n+1)}{\Gamma(n+1)}\to 1 \) for \( A\to\infty \)
In this case indeed summation and limes is not swapable.
Cool investigations!
I'm not entirely sure if it really works, though it seems to. And yeah, if the limit was inside the summation sign it wouldn't go. The integral can be written as the limit of a finite one. This limit sign (not written in the derivation) always stays outside the summation sign.

