Re: Bug in analytical sum

*To*: mathgroup at smc.vnet.net*Subject*: [mg100295] Re: [mg100244] Bug in analytical sum*From*: danl at wolfram.com*Date*: Sun, 31 May 2009 06:35:08 -0400 (EDT)*References*: <200905300057.UAA23355@smc.vnet.net>

> 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

**References**:**Bug in analytical sum***From:*Sebastian Meznaric <meznaric@gmail.com>