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Groups > comp.soft-sys.math.maple > #334

Re: Series Summation Question

From Peter Pein <petsie@dordos.net>
Newsgroups comp.soft-sys.math.maple
Subject Re: Series Summation Question
Date 2012-01-27 07:31 +0100
Organization 1&1 Internet AG
Message-ID <jftgbn$bt$1@online.de> (permalink)
References <1fe7f908-df47-4b54-949f-630a50b0551e@j14g2000vba.googlegroups.com> <87vcnywi1u.fsf@san.rr.com> <39d1ff7b-301a-48ee-99a6-3e4d0df84b7a@o13g2000vbf.googlegroups.com>

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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

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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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