MathGroup Archive 2008

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Question on Sum[] function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86514] Re: Question on Sum[] function
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Thu, 13 Mar 2008 04:28:56 -0500 (EST)
  • References: <fr5du5$o43$1@smc.vnet.net>

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


  • Prev by Date: Re: Path to *.m file
  • Next by Date: Function Works in Notebook NOT Package
  • Previous by thread: Question on Sum[] function
  • Next by thread: Re: Re: Question on Sum[] function