       Re: nested sums

• To: mathgroup at smc.vnet.net
• Subject: [mg125231] Re: nested sums
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Thu, 1 Mar 2012 05:33:03 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201202291226.HAA28458@smc.vnet.net>

```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.
>

```

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