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Re: Problem with a sum
*To*: mathgroup at smc.vnet.net
*Subject*: [mg53893] Re: Problem with a sum
*From*: Maxim <ab_def at prontomail.com>
*Date*: Wed, 2 Feb 2005 06:25:56 -0500 (EST)
*References*: <ctniaq$evd$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On Tue, 1 Feb 2005 09:32:10 +0000 (UTC), <ncc1701zzz at hotmail.com> wrote:
> Hello.
>
> I would like to ask you a question about a sum in a problem I have
> found in Mathematica 5.1.
>
> The sum is the following:
>
> Sum[(k^2 - (1/2))/(k^4 + (1/4)), {k, 1, 1000}]
>
> I have no problems with the sum in that form, but the following one
> doesn't work:
>
>
> s=Sum[(k^2 - (1/2))/(k^4 + (1/4)), {k, 1, m}]
> s /. m->1000
>
>
> It gives a long result with hypergeometric functions. Also, it cannot
> be converted to a number with N[], due to some kind of ComplexInfinity
> problem. FullSimplify doesn't help, neither.
>
>
>
> Also, if I evaluate it to Infinity, I cannot get the value symbolically
> nor numerically, except if I use NSum[], that gives me the right
> result, 1.
>
> I don't know if I'm doing something wrong, it is a bug, a limitation or
> whatever.
>
>
> Thanks a lot for your help.
>
> Best regards.
>
>
Here's what I get in version 5.1:
In[1]:=
s=Sum[(k^2 - (1/2))/(k^4 + (1/4)), {k, 1, m}]
Out[1]=
(1 + m)/((-1 - (1 - I)*m)*(1 + (1 + I)*m)) + HypergeometricPFQ[{1/2 - I/2,
1/2 + I/2, 2}, {3/2 - I/2, 3/2 + I/2}, 1] - HypergeometricPFQ[{3/2 - I/2,
3/2 + I/2, 2}, {5/2 - I/2, 5/2 + I/2}, 1]/5 - HypergeometricPFQ[{2, (3/2 -
I/2) + m, (3/2 + I/2) + m}, {(5/2 - I/2) + m, (5/2 + I/2) + m}, 1]/(5
+ 6*m + 2*m^2) + HypergeometricPFQ[{2, (5/2 - I/2) + m, (5/2 + I/2) + m},
{(7/2 - I/2) + m, (7/2 + I/2) + m}, 1]/(13 + 10*m + 2*m^2) + RootSum[5
+ 16*#1 + 24*#1^2 + 16*#1^3 + 4*#1^4 & , PolyGamma[0, -#1]/(1 + 3*#1
+ 3*#1^2 + #1^3) & ]/8 - RootSum[5 + 16*m + 24*m^2 + 16*m^3 + 4*m^4
+ 16*#1 + 48*m*#1 + 48*m^2*#1 + 16*m^3*#1 + 24*#1^2 + 48*m*#1^2
+ 24*m^2*#1^2 + 16*#1^3 + 16*m*#1^3 + 4*#1^4 &, PolyGamma[0, -#1]/(1 + 3*m
+ 3*m^2 + m^3 + 3*#1 + 6*m*#1 + 3*m^2*#1 + 3*#1^2 + 3*m*#1^2 + #1^3) & ]/8
Mathematica's answer is wrong because HypergeometricPFQ[{1/2 - I/2, 1/2
+ I/2, 2}, {3/2 - I/2, 3/2 + I/2}, 1] is equal to Infinity and we have an
indeterminate expression of the form Infinity - Infinity. However, in this
case we can obtain the correct answer by replacing HypergeometricPFQ[...,
1] with HypergeometricPFQ[..., z] and taking the limit at 1. Mathematica
is unable to compute this limit as well (it will return
DirectedInfinity[0]), but here is a way to make it work:
In[2]:=
Limit[s /. HypergeometricPFQ[La_, Lb_, 1] :>
Series[HypergeometricPFQ[La, Lb, z], {z, 1, 0}] //
Normal,
z -> 1, Direction -> 1]
Out[2]=
(2*m^2)/(1 + 2*m + 2*m^2)
You can verify that it gives the correct result for m = 1000 and also for
m tending to Infinity.
Also the problem can be solved by breaking the summand in two terms:
In[3]:=
Sum[#, {k, m}]& /@ Apart[(k^2 - 1/2)/(k^4 + 1/4)] //
FullSimplify
Out[3]=
(2*m^2)/(1 + 2*m*(1 + m))
Maxim Rytin
m.r at inbox.ru
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