Sum[Sin[n*x]*Sin[n] /. x -> 1, {n, 1, Infinity}] ==

Sum[Sin[n*x]*Sin[n], {n, 1, Infinity}] /. x -> 1

Sum::div: Sum does not converge. >>

Sum::div: Sum does not converge. >>

Sum[Sin[n]^2, {n, 1, Infinity}] == 0

Can anyone explain this? I do not understand why Sum[Sin[n*x]*Sin[n], {n, 1, Infinity}] is automatically simplified to 0 when the series does not converge (unless Mod[x, Pi] == 0 in which case the sum really is zero).

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although this sum is wrong the partial sum given by Mathematica are correct. I have tested this sum with wolfram alpha and it gave me that it diverges so I assume that at v9 they have corrected it …

– Spawn1701D

Apr 21 ’13 at 5:16

1

No, same response from v9.

– fairflow

Apr 21 ’13 at 5:29

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1 Answer

1

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The reason is you hit a distribution.

partialSum = Sum[Sin[n*x]*Sin[n], {n, 1, bigN}];

(* 1/4 Csc[(1 + x)/2] Sin[1/2 (-1 – 2 bigN – x – 2 bigN x)]

-1/4 Csc[(1 – x)/2] Sin[1/2 (-1 – 2 bigN + x + 2 bigN x)] *)

and you can immediately see from here (7th from the top) that in the limit bigN->Infinity you get the sum of two Dirac functions centered at -1 and +1.

So we can conclude that

Sum[Sin[n*x]*Sin[n], {n, 1, Infinity}] = -Pi/2 DiracDelta[x+1] + Pi/2 DiracDelta[x-1]

Another way to check this (not a full proof) is to make the sum convergent :

convSum = Sum[Sin[n*x]*Sin[n] Exp[-eps n], {n, 1, Infinity}, Assumptions :> {eps > 0}];

result = FullSimplify[ComplexExpand[convSum, TargetFunctions -> {Re, Im}]]

(* (Sin[1] Sin[x] Sinh[eps])/

(1 + Cos[2] + Cos[2 x] – 4 Cos[1] Cos[x] Cosh[eps] + Cosh[2 eps]) *)

Plot3D[result, {x, -2, 2}, {eps, 0, 1}, PlotRange -> All]

Limit[result, eps -> 0]

(* 0 *)

but :

Limit[result /. x -> -1, eps -> 0]

(* -Infinity *)

Limit[result /. x -> 1, eps -> 0]

(* Infinity *)