03/24/2010, 05:57 PM
(03/24/2010, 01:11 AM)mike3 Wrote: \( \sum_{n=0}^{\infty} a_n \stackrel{pseudo}{=} \lim_{A \rightarrow \infty} \sum_{n=0}^{\infty} a_n \frac{\gamma(A, n+1)}{\Gamma(n+1)} \)
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!
