Re: Limit

*To*: mathgroup at smc.vnet.net*Subject*: [mg64703] Re: Limit*From*: "Borut Levart" <BoLe79 at gmail.com>*Date*: Tue, 28 Feb 2006 01:49:35 -0500 (EST)*References*: <dtrvmt$lv0$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Helo, First I though your problem deals with continued fraction, but it doesn't really. Perhaps it has an analitical (symbolic) solution, but I don't know it nor do I know how to get it. Perhaps somebody else will enlighten us. I hope so. Anyway, I will show you how to calculate the limit numerically with the neat FoldList construct. To understand the solution, observe what the following line does in Mathematica. FoldList[#2[[1]]/#1 + #2[[2]] &, a6, {{a5, a4}, {a3, a2}, {a1, a0}}] Applied to your example: In[1]:= suum[n_]:=With[{terms=Range[n]}, Fold[#2[[1]]/#1+#2[[2]]&,N[First[terms]],Partition[Reverse[Most[terms]],2]]] In[2]:= suum[11] Out[2]= 1.5422 In[3]:= suum/@Range[1,21,2] Out[3]= {1.,3.,1.28571,1.59459,1.53465,1.5422,1.54143,1.5415,1.54149,1.54149,1.54149} I would say the limit value is approximately 1.54149 You will note that the sum gives the expected figures only for odd numbers. This is because the second argument to Fold must be of even length, but that argument is nothing else than the input list clipped by the last element (which must be sumed up first; the expression is summing up inside out, - this is way the input list is rotated before being passed as the second argument to Fold). Do you like the solution? Bye, Borut Levart, Slovenia