Series Summation Question

[clashton]

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?

[Axel Vogt]

Using Maple 15:

Sum(q^(-6+4*n)/(1-q^(-5+4*n)),n = 0 .. infinity);
eval(%, q=0.1);
evalf(%);
-21.10120010

[Joe Riel]

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

[clashton]

On Jan 26, 11:21 am, Joe Riel <j...@san.rr.com> wrote:

Maple wins this round.

[Peter Pein]

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

[Peter Pein]

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

[clashton]

Thanks to everyone.

I am happy to see that there was a way of doing it in Mathematica, as
I do not have Maple.

[Axel Vogt]

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

>> 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

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)

[Peter Pein]

Am 28.01.2012 20:21, schrieb Axel Vogt:
> 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)

yes, see: http://reference.wolfram.com/mathematica/ref/QPolyGamma.html
and/or
http://mathworld.wolfram.com/q-PolygammaFunction.html