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I am trying to evaluate a infinite series: $$ 2 \sum _{l=1}^{\infty } \Re\left(n^{\frac{2 i \pi l}{\log (q)}} \Gamma \left(\alpha -\frac{2 i l \pi }{\log (q)}\right)\right) $$ With the code:

delta[\[Alpha]_, q_, n_] = 
 2 Sum[Re[Gamma[\[Alpha] - I 2 l \[Pi]/Log[q]] Exp[
      I 2 l \[Pi] Log[n]/Log[q]]], {l, 1, Infinity}]
delta[1,10,10]
N[%]

But Mathematica complaints:


NSum::nsnum: Summand (or its derivative) ((0. +3.14159 I) 2.^(1. +(0. +2.72875 I) l) 5.^((0. +2.72875 I) l) Gamma[1. -(0. +2.72875 I) l]-(0. +1.36438 I) 2.^(1. +(>) l) > Gamma[1. -(0. +>) l] PolyGamma[0.,1. -(0. +2.72875 I) l]) > is not numerical at point l = 16.

Please help me to evaluate it or explain the problem, Thanks!

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Using Mathematica 10.0.2.0, I do not receive the error message you did. However, Sum returns unevaluated, which happens when Sum cannot do the summation. Since you are seeking a numerical answer, I suggest that you use a large upper bound on your Sum instead of Infinity. For instance,

delta[\[Alpha]_, q_, n_] := 2 Sum[Re[Gamma[\[Alpha] - I 2 l \[Pi]/Log[q]] 
    Exp[I 2 l \[Pi] Log[n]/Log[q]]], {l, 1, 100}];
N[delta[1, 10, 10]]
(* 0.0818184 *)

Incidentally, I tried NSum instead, but it did not return an answer in a reasonable time.

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  • $\begingroup$ Thank you for trying the code. Maybe I am too greed to let Mathematica to compute the infinite sum. You're right, I am trying a bigger upper bound now. By the way, I am using Mathematica 9.0.1.0 Thanks very much! $\endgroup$ – robit Jan 20 '15 at 8:29
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Clear[delta]

delta[\[Alpha]_?NumericQ, q_?NumericQ, n_?NumericQ] :=
  Module[
   {z = I 2.0` l \[Pi]/Log[q]},
   Re[2 NSum[Gamma[\[Alpha] - z] n^z,
      {l, 1, Infinity}]]];

delta[1, 10, 10]

0.0818184

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  • $\begingroup$ It does work! And produce the answer in a reasonable time,While The code of @bbgodfrey did not. But why is the use of Module accelerate the calculation of Nsum? Because the compact expression with local variable z? $\endgroup$ – robit Jan 21 '15 at 6:24
  • $\begingroup$ The compact form (reuse of sub calculation of local variable z) helps to speed the calculation up; however, the major cause of the speed up is doing the calculation with approximate real numbers rather than exact numbers and using NSum rather than Sum. $\endgroup$ – Bob Hanlon Jan 21 '15 at 13:39

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