Re: Question on Sum[] function

*To*: mathgroup at smc.vnet.net*Subject*: [mg86631] Re: Question on Sum[] function*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sat, 15 Mar 2008 17:43:10 -0500 (EST)*References*: <fr5du5$o43$1@smc.vnet.net> <fras4i$18m$1@smc.vnet.net>

On Thu, 13 Mar 2008, David W.Cantrell wrote: > sigmundv at gmail.com wrote: >> Dear group members, >> >> If I evaluate Sum[Log[x^n],{n,1,Infinity}] I get -Log[x]/12 as an >> answer, but if I plug any other value than -1 or +1 in, Mathematica >> tells me that the series is divergent; for x=-1 the sum is -I*Pi/12 >> and for x=1 the sum is 0, of course. >> >> If we restrict ourselves to real numbers, I would say that the series >> is only meaningful for x>0, because for x<0, Log[x^n] is not defined >> for odd n. For x>0, we write Sum[Log[x^n],{n,1,Infinity}] as >> Sum[n*Log[x],{n,1,Infinity}], and clearly this series is only >> convergent for x=1, with sum 0. >> >> Well, my actual question was how to interpret the closed form >> expression that Mathematica gives for the sum of the afore-mentioned >> series. Mathematica ought to return to me some condition on x, because >> Sum[Log[x^n],{n,1,Infinity}] == -Log[x]/12 is not true for all real, >> or complex, x. >> >> I hope that you can shed some light on this. > > Since nobody has answered yet... > > For simplicity, let's assume that x > 0, so that, as you said, we may > "write Sum[Log[x^n],{n,1,Infinity}] as Sum[n*Log[x],{n,1,Infinity}]". > If x is not 1, then the latter sum is formally > > Log[x] * Sum[n,{n,1,Infinity}] (*) > > Now of course Sum[n,{n,1,Infinity}] diverges to Infinity. But there is a > sense in which it equals -1/12: > > Consider > > In[6]:= Sum[n^p, {n, 1, Infinity}] > > Out[6]= Zeta[-p] > > Now of course if p is 1, the sum in In[6] is Sum[n,{n,1,Infinity}]. > Replacing p by 1 in Out[6], we get > > In[7]:= Zeta[-1] > > Out[7]= -1/12 > > "Therefore", Sum[n,{n,1,Infinity}] equals -1/12. Substituting that result > into (*), we get > > -Log[x]/12 > > which is the result which Mathematica had given for your original sum. > > You're welcome to think of what I said as being rather absurd. Yet, I > suspect that what I said is actually related to the reason that > Mathematica gave -Log[x]/12 for your sum. > > David > Hello Sigmund and David, The behavior of Sum in the above example is unintentional. As noted by David, the factor of (-1/12) in the answer comes from Zeta[-1]. The expected behavior for this sum would be for it to return unevaluated, except when x = 1. Thank your for the example and for your comments on the problem, which clearly needs to be fixed. Sincerely, Devendra Kapadia, Wolfram Research, Inc.