From another question I tried to implement something like:

Sum[r^(1 - l[n]) DD[n], {n, Complement[Range[1, Infinity], {4}]}]

but this does not work because Range does not take infinite limits.

Is there another way to exclude a certain numeric value for the index from an infinite sum?

  • $\begingroup$ Sum[r^(1 - l[n]) DD[n], {n, 1, 3}] + Sum[r^(1 - l[n]) DD[n], {n, 5, Infinity}] $\endgroup$ – Bob Hanlon Mar 5 '15 at 18:12
  • $\begingroup$ Well, this is not very practical if I have many excluded values $\endgroup$ – usumdelphini Mar 5 '15 at 18:35

Since you know your exclusions beforehand, you can proceed like this (for any number of exclusions!):

excl={4,9,14}; (* just an example of more than one exclusion! *)
Sum[r^(1-l[n]) DD[n],{n, Select[Range[1, Max@excl],!MemberQ[excl,#]&]}]+
  Sum[r^(1-l[n]) DD[n],{n,Max@excl+1,Infinity}]

This will filter out all your exclusions in one run.

It will also deal with exclusions you make to avoid the summed expression to become invalid at those points.

| improve this answer | |

An straightforward way to put exclusions into sums is by using KroneckerDelta:

excl = {4, 9, 14};
  Times @@ (1 - Thread[KroneckerDelta[n, excl]]) r^(1 - l[n]) DD[n], 
 {n, 1, Infinity}

$$\sum _{n=1}^{\infty } \text{DD}(n) (1-\delta _{4,n}) (1-\delta _{9,n}) (1-\delta _{14,n}) r^{1-l(n)}$$

To show that this also works when the summand would otherwise be undefined, here is an example:

DD[n_] := 1/(n - 9)^2

 Times @@ (1 - Thread[KroneckerDelta[n, excl]])  DD[n], {n, 1, 

(* ==> (1021301 + 117600 Pi^2)/705600 *)

So the divergent term never gets evaluated because the KroneckerDelta sets it to zero first.

| improve this answer | |
  • $\begingroup$ Nice! One drawback, though: It does not deal with the case, that exclusions are made because of the summed expression is not defined at those points. $\endgroup$ – Jinxed Mar 5 '15 at 22:24
  • $\begingroup$ @Jinxed Actually, it does. I'll give an example in my answer. $\endgroup$ – Jens Mar 5 '15 at 22:49

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