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Is there any way to force Mathematica to come up with the closed form to

Sum[1/(i^18 (i^2 + j^2)), {i, 1, Infinity}, {j, 1, Infinity}]

or

Sum[1/(i^4 (i^2 + j^2)), {i, 1, Infinity}, {j, 1, Infinity}]

? The former series definitely has a closed form, whereas the 2nd one might have or not.

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I conjecture that all double series having at denominator the form $i^{4k+2} (i^2 + j^2)$ could be expressed in terms of $\pi$ and zeta function. –  Chris's sis Jul 27 '13 at 7:37

2 Answers 2

up vote 7 down vote accepted

I like to use a little trick.

Apart[1/(i^18 (i^2 + j^2)), i]
(* 1/(i^2 j^18) - 1/(i^4 j^16) + 1/(i^6 j^14) - 1/(i^8 j^12) + 
   1/(i^10 j^10) - 1/(i^12 j^8) + 1/(i^14 j^6) - 1/(i^16 j^4) + 1/(i^18 j^2) 
   - 1/(j^18 (i^2 + j^2)) *)

All the terms except the last one are easy. When summed upon, the last term equals our original series due to symmetry, therefore we can just sum the easy bits and then divide by 2.

aux = Sum[ 1/(i^2 j^18) - 1/(i^4 j^16) + 1/(i^6 j^14) - 1/(i^8 j^12) + 
           1/(i^10 j^10) - 1/(i^12 j^8) + 1/(i^14 j^6) - 1/(i^16 j^4) + 
           1/(i^18 j^2), {i, 1, Infinity}] ;

result = 1/2 Sum[aux, {j, 1, Infinity}]
(* (584887 \[Pi]^20)/4764136114237500 *)

Check :

N[result]
(* 1.07668 *)

NSum[1/(i^18 (i^2 + j^2)), {i, 1, Infinity}, {j, 1, Infinity}]
(* 1.07668 *)

I think one can use the same way for the other sum.

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1  
Very nice and useful this trick. (+1) –  Chris's sis Jul 27 '13 at 8:01
    
The other sum doesn't work this way. –  Chris's sis Jul 27 '13 at 8:06

First problem

Break the sum into 2 parts:

Clear[m]; 
s1 = Sum[1/(i^18*(i^2 + j^2)), {j, 1, m}]; 
s1 = Limit[s1, m -> Infinity]

Out[49]= (-1 + i*Pi*Coth[i*Pi])/(2*i^20)

Now the second part, Mathematica can't sum. i.e s2 = Sum[s1, {i, 1, m}], but we can numerically look at it for larger and larger m values...

data = Table[{m, NSum[s1, {i, 1, m}]}, {m, 1, 100}];
ListPlot[data, Joined -> True, Mesh -> All, PlotRange -> All]

Mathematica graphics

data[[-1, 2]]
Out[53]= 1.07668

Second problem

 Clear[m]
 s1 = Sum[1/(i^4*(i^2 + j^2)), {j, 1, m}]; 
 s1 = Limit[s1, m -> Infinity]

 Out[28]= (-1 + i*Pi*Coth[i*Pi])/(2*i^6)

again, s2 = Sum[s1, {i, 1, Infinity}] does not sum. So try numerical

data = Table[{m, NSum[s1, {i, 1, m}]}, {m, 1, 100}];
ListPlot[data, Joined -> True, Mesh -> All, PlotRange -> All]

Mathematica graphics

 data[[-1, 2]]
 Out[46]= 1.12601
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Thanks for answer. (+1) –  Chris's sis Jul 27 '13 at 8:01

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