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From: Chris H. Fleming on 2 Mar 2010 04:14
Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]

Sum does not converge.

NSum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]

0.206647


Fortunately I know how to do this sum by hand, but Mathematica can
usually handle these Harmonic number functions pretty well.

Does anyone know a way of massaging this into a form Mathematica can
digest?

From: Bob Hanlon on 2 Mar 2010 07:58


$Version

7.0 for Mac OS X x86 (64-bit) (February 19, 2009)

s = Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, Infinity}]

(-(1/2))*RootSum[#1^4 + 4*#1^3 +
11*#1^2 + 14*#1 + 10 & ,
PolyGamma[0, -#1]/(2*#1^2 +
4*#1 + 7) & ]

s // N // Chop

0.206647

s // ToRadicals // Simplify

(1/6)*(-PolyGamma[0, 1 - I] -
PolyGamma[0, 1 + I] +
PolyGamma[0, 1 - 2*I] +
PolyGamma[0, 1 + 2*I])

% // N // Chop

0.206647

s // ToRadicals // FullSimplify

(1/6)*(-HarmonicNumber[-I] -
HarmonicNumber[I] +
HarmonicNumber[-2*I] +
HarmonicNumber[2*I])

% // N // Chop

0.206647

NSum[k/((k^2 + 1) (k^2 + 4)), {k, 1, Infinity}]

0.206647


Bob Hanlon

---- "Chris H. Fleming" <chris_h_fleming(a)yahoo.com> wrote:

=============
Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]

Sum does not converge.

NSum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]

0.206647


Fortunately I know how to do this sum by hand, but Mathematica can
usually handle these Harmonic number functions pretty well.

Does anyone know a way of massaging this into a form Mathematica can
digest?



From: Fabrice P. Laussy on 2 Mar 2010 07:59
Chris H. Fleming wrote:
> Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]
>
> Sum does not converge.
[...]
> Fortunately I know how to do this sum by hand, but Mathematica can
> usually handle these Harmonic number functions pretty well.

What result do you get by hand?

What Mathematica version to you have? V�7.0.0 returns the result as the
roots of a quartic polynomial over the polygamma function.

From: Chris H. Fleming on 2 Mar 2010 08:00
On Mar 2, 4:14 am, "Chris H. Fleming" <chris_h_flem...(a)yahoo.com>
wrote:
> Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]
>
> Sum does not converge.
>
> NSum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]
>
> 0.206647
>
> Fortunately I know how to do this sum by hand, but Mathematica can
> usually handle these Harmonic number functions pretty well.
>
> Does anyone know a way of massaging this into a form Mathematica can
> digest?


Everyone else was getting the right answer, so I went back through my
file to look for the culprit.
I have found the problem.

$Assumptions = {k > 0};
Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]

Sum does not converge.

I had required k>0 to get to this point (convergent integrals) but
didn't imagine that it would have any effect on this sum.

Strangely a numerator of k^0 or k^2 will work, just not k^1.

From: Patrick Scheibe on 2 Mar 2010 08:00
Hi,

no problems here (OSX, Math 7)

In[5]:= Sum[k/((k^2+1) (k^2+4)),{k,1,\[Infinity]}]//FullSimplify

Out[5]= 1/6 (-HarmonicNumber[-I]-HarmonicNumber[I]+HarmonicNumber[-2 I]
+HarmonicNumber[2 I])

In[6]:= %//N
Out[6]= 0.206647+0. I

Cheers
Patrick

Am Mar 2, 2010 um 9:34 AM schrieb Chris H. Fleming:

> Sum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]
>
> Sum does not converge.
>
> NSum[k/((k^2 + 1) (k^2 + 4)), {k, 1, \[Infinity]}]
>
> 0.206647
>
>
> Fortunately I know how to do this sum by hand, but Mathematica can
> usually handle these Harmonic number functions pretty well.
>
> Does anyone know a way of massaging this into a form Mathematica can
> digest?
>


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