MMA differently handles division by zero. How to control this behavior?
In this case it delivers the expected error:
i/(i + k) /. {i -> 0, k -> 0}
(* Indeterminate *)
but if you calculate the same using Sum
Sum[i/(i + k), {i, 0, 0}] /. k -> 0 or Sum[0/k, {i, 0, 0}] /. k -> 0
(* 0 *)
the result is 0 presumably because MMA internally uses the limit:
Limit[0/(0 + k), k -> 0]
(* 0 *)
Tracethe evaluation you will see that theSumis evaluated symbolically prior to the replacement. The symbolicSumis zero and there is nokto be replaced. Also,Inactive[Sum][i/(i + k), {i, 0, 0}] /. k -> 0 // Activateevaluates to1since Mathematica always immediately simplifiessymbol/symbolto1$\endgroup$symbol/symbolin the first case? How can the symbol reduction in the sum be switched off? $\endgroup$Traceand you will see that both variables are first replaced by zero resulting in0/(0+0)so there is nosymbol/symbol$\endgroup$Piecewiseor equivalent structure. $\endgroup$Sum[]is evaluated in your code before the replacementk -> 0(which is in no way connected with limits). The sum becomes0, because0/kevaluates to0wheni = 0. Then all instances ofkthat are left (which is none) are replaced by0.Block[{k = 0}, Sum[i/(i + k), {i, 0, 0}]]in effect haskandievaluated simultaneously each time a term is computed.Sum[Evaluate[i/(i + k) /. k -> 0], {i, 0, 0}]haskevaluated first, before the sum is computed.Trace[]will show these steps. $\endgroup$