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July 1st, 2012, 06:47 PM  #1 
Member Joined: Jan 2012 Posts: 72 Thanks: 0  Convergence of a recurrence relation
Given . Use a calculator to determine the behaviour of the sequence for each of the cases I had no problems doing and . For , the series increases and converges to 0.619 while for , the series decreases and converges to 0.619. However, when solving for , all i got was error on the calculator screen. 
July 1st, 2012, 07:25 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Convergence of a recurrence relation
I observed the same behavior you did, and for the series diverged. For it increased and converged to 0.619061286736 For it decreased and converged to 0.619061286736 I tried finding the closed form via Newton's method to create a first order IVP, but wound up with an integral without an antiderivative expressible in elementary terms. I found by setting that the value the series converged to in the first two cases is one of the solutions to: The other solution is about 1.5121345516578424739. Here is a plot: [attachment=0:5ucqk92t]convergence.jpg[/attachment:5ucqk92t] The larger root seems to be the division between convergence and divergence for the given recurrence relation when used as the starting value. 
July 1st, 2012, 08:02 PM  #3 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Convergence of a recurrence relation
Solving for fixed points, where W(z) is the Lambert W function, which can be assigned two values there, 0.619061 and 1.51213. The first is stable and the second is unstable; any starting value greater than 1.51213 will result in divergence. 
July 1st, 2012, 08:10 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Convergence of a recurrence relation
In fairness to [color=#0040FF]aswoods[/color], I was editing my post to include essentially the same information (albeit not as succinctly) in his post at the same time he was posting. 

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