Re: nested sums

*To*: mathgroup at smc.vnet.net*Subject*: [mg125254] Re: nested sums*From*: Severin Pošta <severin at km1.fjfi.cvut.cz>*Date*: Fri, 2 Mar 2012 07:49:45 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201202291226.HAA28458@smc.vnet.net> <jinjbt$ctp$1@smc.vnet.net>

Sure, but that's not what I am asking for, I wanted to ask if such check can be somehow performed automatically? Thanks, S. On 3/1/2012 11:34 AM, Bob Hanlon wrote: > Sum[x^a y^b z^c, {a, 0, Infinity}, {b, 0, a}, {c, 0, a - b}] > > -(1/((-1 + x)*(-1 + x*y)*(-1 + x*z))) > > > Bob Hanlon > > 2012/2/29 Severin Po=C5=A1ta<severin at km1.fjfi.cvut.cz>: >> I have noticed that >> >> Sum[x^a y^b z^c, {a, 0, Infinity}, {b, 0, Infinity}, {c, 0, a - b}] >> >> produces "wrong" result >> >> (-y + z - y z + x y z)/((-1 + x) (-1 + y) (y - z) (-1 + x z)) >> >> because a-b is not checked to be nonnegative in the nested sums. This is >> probably by design (?). >> I wonder if I can achieve such check to be perfomed. >> >> This does not work: >> >> Sum[x^a y^b z^c If[a - b>= 0, 1, 0], {a, 0, Infinity}, {b, 0, >> =C2 Infinity}, {c, 0, a - b}] >> >> This also does not work: >> >> Sum[x^a y^b z^c Piecewise[{{1, a - b>= 0}}], {a, 0, Infinity}, {b, 0, >> =C2 =C2 Infinity}, {c, 0, a - b}] >> >> >> It is also interesting that Sum[...,{},{}] is not equivalent to >> Sum[Sum[...,{}],{}]. The following produces correct result: >> >> s= Sum[x^a y^b z^c Piecewise[{{1, a - b>= 0}}], {c, 0, a - b}]; >> s1=Sum[s, {b, 0, Infinity}]; >> Sum[s1,{a, 0, Infinity}] >> >> >> -(1/((-1 + x) (-1 + x y) (-1 + x z))) >> >> >> Interesting. >> >> S. >> >