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Prove a formula for a known convergent series with Resolve?

Posted 5 months ago
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I ran across this behaviour that I do not understand. I wanted to prove a formula for a known convergent series:

$\sum_{k=-\infty}^\infty \frac{1}{(2k+1)(2q-2k-1)} = -\frac{\pi^2}{4} \delta_{q,0}\quad$ for $\quad q\in\mathbb{Z}$.

However, Mathematica says this is wrong:

Resolve[ForAll[q, Sum[1/((2  k + 1) (2  q - 2  k - 1)) == -(\[Pi]^2/4)  KroneckerDelta[q, 0], {k,-Infinity,Infinity}], Integers]

returns False. Moreover, if the values for $q$ are restricted, i.e. if I look separately at the cases $q=0$ and $q\neq 0$, I get a True result: The two commands

Resolve[ForAll[q, q == 0, Sum[1/((2  k + 1) (2  q - 2  k - 1)) == -(\[Pi]^2/4)  KroneckerDelta[q, 0], {k,-Infinity,Infinity}], Integers]

Resolve[ForAll[q, q != 0, Sum[1/((2  k + 1) (2  q - 2  k - 1)) == -(\[Pi]^2/4)  KroneckerDelta[q, 0], {k,-Infinity,Infinity}], Integers]

both evaluate to True. Logically, this is a contradiction, so my question is whether I misunderstood the function of these commands or this is a bug. Thanks in advance for clarification! (Using Mathematica 10.0)

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Hi Julian,

I guess the problem is that for the sum first the general case is evaluated, and this gives:

Sum[1/((2 k + 1) (2 q - 2 k - 1)), {k, -Infinity, Infinity}]
(*  Out:    -((\[Pi] Tan[\[Pi] q])/(4 q))   *)

So your first line reads:

Resolve[ForAll[q, -((\[Pi] Tan[\[Pi] q])/(4 q)) == -(\[Pi]^2/4) KroneckerDelta[q, 0]], Integers]

and this evaluates to False, because in the case $q=0$ we have:

(-((\[Pi] Tan[\[Pi] q])/(4 q)) == -(\[Pi]^2/4)) /. q -> 0
(*  Out:    Indeterminate\[Equal]-(\[Pi]^2/4)   *)

In this case you need the limit:

Limit[-((\[Pi] Tan[\[Pi] q])/(4 q)), q -> 0]

If you restrict the value of $q$ before the sum gets evaluated (you second option above) everything works.

Hope that helps, regards -- Henrik

Thanks for the insight!

This explains the result, but I am somewhat disappointed that the general case yields

-((\[Pi] Tan[\[Pi] q])/(4 q))

and not the continuous continuation of this expression.

Best regards, Julian

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