> 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?

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?

> 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

> 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.

> 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

> Am 27.01.2012 07:31, schrieb Peter Pein:

> >>> 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

>> 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

> On 27.01.2012 07:55, Peter Pein wrote:
> ....

>> 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

> I am not aware that Maple has a command to find a
> condition for convergence.

> And it does not provide a symbolic solution (where
> I guess the above cryptic command does just that)