Take the following code:

      ((-1)^(1 + q1 + q2) 2^(-2 q1 - 2 q2) q2 Theta^(1 + 2 q1 + 2 q2))/((1 + 2 q1)! (2 q2)!), 
      {q1, 0, Infinity}], 
    {q2, 0, Infinity}]]

    ((-1)^(1 + q1 + q2) 2^(-2 q1 - 2 q2) q2 Theta^(1 + 2 q1 + 2 q2))/((1 + 2 q1)! (2 q2)!), 
    {q1, 0, Infinity}, {q2, 0, Infinity}]]

I am basically summing the same coefficients, but in two different ways: one with two summations, the other with a summation over two indices. The results are the following:

1/2 Theta Sin[Theta/2]^2

-(1/4) Theta Cos[Theta]

As you can see they dont match. Why is it so?

To give some interpretation, it is actually the Taylor coefficient of sin(theta/2) * q1) * Taylor_coefficient(cos(theta/2)) * q2 * (-2 q2) that I integrate. Said differently, I am summing

-2 q2 Coeff[Sin[q1]] Coeff[Cos[q2]]

over q1 and q2 :

  • $\begingroup$ Last point is not clear. Can you please write more precisely how you get the series, either in mathematical form or a respective derivation with MA. The equation in orange box has no Theta. $\endgroup$
    – yarchik
    Dec 3, 2019 at 13:59
  • $\begingroup$ You could substitute numerical values to find which gives the correct answer. Remember that it may not be mathematically legitimate to interchange the 2 limiting processes. $\endgroup$
    – mikado
    Dec 3, 2019 at 17:57

1 Answer 1


Not an answer, just a long comment...

One can simplify this case to this:

expr = ( 2^(q1 + q2) q1)/((1 + 2 q1)! ( 2 q2)!);

Sum[expr, {q1, 0, Infinity}, {q2, 0, Infinity}, Method -> "IteratedSummation"] -
Sum[expr, {q1, 0, Infinity}, {q2, 0, Infinity}] // FullSimplify

(* 1/4 *)

Which more clearly specifies the discrepancy in two methods.

  • $\begingroup$ To understand: is it then a a problem of interversion of two series (thus a mathematical problem) behind ? I tried to look at what means "iterated univariate summation" but I didn't find something interesting yet. $\endgroup$
    – StarBucK
    Dec 3, 2019 at 16:39
  • $\begingroup$ @StarBucK I suspect it's a bug in Automatic method Wolfram uses to calculate this double sum. IteratedSummation is a method of just symbolically summing one sum and then another, like in your question Sum[Sum[...],...] $\endgroup$
    – swish
    Dec 3, 2019 at 16:45

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