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MathGroup Archive 2006

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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


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