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Groups > comp.soft-sys.math.maple > #335

Re: Series Summation Question

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Am 27.01.2012 07:31, schrieb Peter Pein:
> Am 27.01.2012 01:45, schrieb clashton:
>> On Jan 26, 11:21 am, Joe Riel<j...@san.rr.com> wrote:
>>> clashton<clash...@gmail.com> writes:
>>>> I am a little curious about whether Maple can evaluate a certain
>>>> infinite sum that fails with Mathematica (I do not have Maple so
>>>> cannot test it myself).
>>>
>>>> Here is the sum in Mathematica:
>>>> Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
>>>> This leads to and error,
>>>> but
>>>> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q -> 0.1)
>>>> +
>>>> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q -> 0.1})
>>>> gives the correct answer, which I find a little strange.
>>>
>>>> How does Maple perform on the original sum?
>>>
>>> Summing numerically (and guessing at the meaning of the Mathematica):
>>>
>>> (**) S := Sum(q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), n=0..infinity);
>>> infinity
>>> ----- (-6 + 4 n)
>>> \ q
>>> S := ) ---------------
>>> / (-5 + 4 n)
>>> ----- 1 - q
>>> n = 0
>>>
>>> (**) evalf(eval(S,q=1/10));
>>> -21.10120010
>>>
>>> --
>>> Joe Riel
>>
>> Maple wins this round.
>
> Well, if you are interested in the approximation of the sum's value for
> q=1/10, you should have used NSum in Mathematica:
>
> NSum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)) /. q->1/10, {n, 0, Infinity},
> WorkingPrecision->20]
>
> -21.101200102001001100
>
> or - using your workaround - get
>
> 6/(5*(-1 + q)) + (1 + 2*q + 3*q^2 + 4*q^3)/(5*(1 + q + q^2 + q^3 + q^4))
> + (-2*Log[q^4] + Log[1 - q^4] + QPolyGamma[0, 2 - Log[q^5]/Log[q^4],
> q^4])/(q*Log[q^4])
>
> as (hopefully) exact value.
>
> Cheers, Peter

Sorry for posting too fast. One gets the result in a more simple form by 
doing sth. more complicated:

In[1]:= f[q_] = q^(-6 + 4*n)/(1 - q^(-5 + 4*n));
assum = SumConvergence[f[q], n]
s[q_] = Together[Subtract @@ (Limit[Sum[f[q], n], n -> #1, Assumptions 
-> assum] & ) /@ {Infinity, 0}]
N[s[1/10]]


Out[2]= q != 0 && Abs[q]^4 < 1
Out[3]= (Log[1 - q^4] + QPolyGamma[0, -(Log[q^5]/Log[q^4]), 
q^4])/(q*Log[q^4])
Out[4]= -21.1012

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Thread

Series Summation Question clashton <clashton@gmail.com> - 2012-01-25 22:46 -0800
  Re: Series Summation Question Axel Vogt <&noreply@axelvogt.de> - 2012-01-26 19:41 +0100
  Re: Series Summation Question Joe Riel <joer@san.rr.com> - 2012-01-26 08:21 -0800
    Re: Series Summation Question clashton <clashton@gmail.com> - 2012-01-26 16:45 -0800
      Re: Series Summation Question Peter Pein <petsie@dordos.net> - 2012-01-27 07:31 +0100
        Re: Series Summation Question Peter Pein <petsie@dordos.net> - 2012-01-27 07:55 +0100
          Re: Series Summation Question clashton <clashton@gmail.com> - 2012-01-28 07:59 -0800
          Re: Series Summation Question Axel Vogt <&noreply@axelvogt.de> - 2012-01-28 20:21 +0100
            Re: Series Summation Question Peter Pein <petsie@dordos.net> - 2012-02-02 12:21 +0100

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