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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