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Re: Question on Sum[] function

On Thu, 13 Mar 2008, David W.Cantrell wrote:
> sigmundv at 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.


Devendra Kapadia,
Wolfram Research, Inc.

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