       Re: Bug in Sum?

• To: mathgroup at smc.vnet.net
• Subject: [mg120035] Re: Bug in Sum?
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Wed, 6 Jul 2011 08:28:56 -0400 (EDT)

```You can also try summing the parts

Sum[#, {n, 1, M}] & /@
Apart[1/(n (1 + n)) 1/((1 - x)^4), n] //
FullSimplify

M/((1 + M) (-1 + x)^4)

Bob Hanlon

---- Dario <dario.benedetti at aei.mpg.de> wrote:

=============
Thanks Phil, I tried both with version 7 (Mac and Windows) and 8 (Mac) and it doesn't work.

Bob, thanks for the suggestion, but the real problem is more complicated, I get some long and complicated output from some previous stuff, and as I couldn't sum it I discovered this bug.

> Works in version 6.0
>
> In:= Sum[1/(n (1 + n)) 1/((1 - x)^4), {n, 1, M}]
> \$Version
>
> Out= M/((1 + M) (1 - x)^4)
>
> Out= "6.0 for Microsoft Windows (32-bit) (June 19,
> 2007)"
>
> On Thu, Jun 30, 2011 at 6:29 AM, Bob Hanlon
> <hanlonr at cox.net> wrote:
>
> >
> > For a workaround, generalize the problem
> >
> > Sum[1/(n (1 + n)) 1/((1 - x)^r), {n, 1, M}]
> >
> > M/((1 - x)^r*(1 + M))
> >
> > % /. r -> 4
> >
> > M/((1 + M)*(1 - x)^4)
> >
> >
> > Bob Hanlon
> >
> > ---- Dario <dario.benedetti at aei.mpg.de> wrote:
> >
> > =============
> > I am very puzzled by the fact that Mathematica does
> not evaluate the
> > following Sum (it keeps running endlessly):
> >
> > Sum[1/(n  (1 + n)) 1/((1 - x)^4) , {n, 1, M}]
> >
> > where x and M are some undefined variables,
> >
> > but it does evaluate
> >
> > Sum[1/(n  (1 + n)) 1/((1 - x)^2) , {n, 1, M}]
> >
> > where I have only changed the power of (1-x), which
> should not matter as it
> > must factor out!
> >
> > Any help on this please?
> >

```

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