       Re: Bug in Sum?

• To: mathgroup at smc.vnet.net
• Subject: [mg119978] Re: Bug in Sum?
• From: Dana DeLouis <dana01 at me.com>
• Date: Sun, 3 Jul 2011 04:10:51 -0400 (EDT)

```Hi.  If x had an assumed value of 1, then the equation would error with
Div by zero.  However, no combination of assumptions, or Methods, appear
to overcome this.
If we change (1-x) to a single variable, then it appears to work.

equ=1/(n (1+n)) 1/((x)^4);

Sum[equ,{n,m}]
m/((1+m) x^4)

% /. x->(1-x)
m / ((1+m) (1-x)^4)

I'm not sure why x works, and not 1-x.

Even Wolfram Alpha has a hard time with this.

WolframAlpha["Sum (1/(n*(1 + n)*(x)^4)), n=1 to m=94]

<< Nice Output with Graph...>>

However, (1-x) causes a problem...

WolframAlpha["Sum (1/(n*(1 + n)*(1-x)^4)), n=1 to m=94]

.. Wolfram Alpha doesn't know how to interpret your input

Wow.  I don't see the problem.  Changing (x) to (1-x), and even Wolfram Alpha doesn't even know how to interpret this.

Not sure, but it sounds like a logic bug somewhere, unless there's a
mathematical reason for this behavior.
Anyone??

= = = = = = = = = =
HTH  : >)
Dana DeLouis
\$Version
8.0 for Mac OS X x86 (64-bit) (November 6, 2010)

On Jun 29, 5:42 pm, Dario <dario.benede... 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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