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Re: Bug in analytical sum

  • To: mathgroup at
  • Subject: [mg100295] Re: [mg100244] Bug in analytical sum
  • From: danl at
  • Date: Sun, 31 May 2009 06:35:08 -0400 (EDT)
  • References: <>

> Consider the following sum
> f[k_] := Sum[n! (-I \[Alpha])^n Cos[\[Phi]/2]^n \[Alpha]^n Sin[\[Phi]/
> 2]^n, {n,
>     0, k}]/Sum[n! \[Alpha]^(2 n) Cos[\[Phi]/2]^(2 n), {n, 0, k}]
> I am interested in taking alpha and k to infinity. Now clearly for
> finite k this is just a rational function in alpha. So if we want to
> take alpha to infinity we should get
> (-i Tan[\[Phi]/2])^k.
> But try this in Mathematica Limit[f[k], \[Alpha] -> \[Infinity]] and
> you will get I Cot[\[Phi]/2]. This is Mathematica 7.0.0.
> The reason seems to be that Mathematica evaluates the sum first and
> obtains a fraction consisting of the incomplete gamma functions and Ei
> integrals. It seems that the limits of those functions are not taken
> properly.

Offhand I do not know what is the correct result. But I can say that Limit
will have trouble managing branch cuts, if not given suitable assumptions
(and it may have trouble anyway...). So you might instead do:

In[18]:= Limit[f[k], \[Alpha] -> Infinity, Assumptions ->
     {k > 0, Element[\[Phi], Reals]}]

Out[18]= (-(1/2))^k/(Cos[\[Phi]/2]^(2*k)*((-I)*Csc[\[Phi]])^k)

Daniel Lichtblau
Wolfram Research

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