12/11/2009, 08:16 AM
(12/11/2009, 06:27 AM)Ansus Wrote:(12/10/2009, 08:48 PM)mike3 Wrote: \( \sum_{n=0}^{x-1} f(n) = f(0) + \sum_{n=1}^{\infty} f(n) - f(n + x - 1) \).
Or
\( \sum_{n=0}^{x-1} f(n) =- f(0) + \sum_{n=0}^{\infty} f(n) - f(n + x) \).
\( \sum_{n=0}^{x-1} f(n) =f(0) - \sum_{n=0}^{\infty} f(n + x)-f(n) \).
But this only works for functions which go to 0 at infinity.
If it does not go to 0 but to a fixpoint: can we subtract that constant/fixpoint and compensate for that by introducing + zeta(0)*fixpoint ? (In some fiddling in other examples weeks ago such a compensation seemed to be meaningful, don't have it at hand at the moment, though)
Gottfried
Gottfried Helms, Kassel

