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MathGroup Archive 2006

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Re: Limit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64692] Re: Limit
  • From: "Scout" <Scout at nodomain.com>
  • Date: Tue, 28 Feb 2006 01:49:24 -0500 (EST)
  • References: <dtrvmt$lv0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Sinan Kapçak" <sinankapcak at yahoo.com>
>i want to find the value of the limit
>
> 1+(2/(3+(4/(5+(6/(7+...
>
> how can i do that with Mathematica?
>
> Tnx..
>
Hi Sinan,
you could try to solve the recurrence equation that defines your continued 
fraction:
i.e.
f[x_]:= x+(x+1) / f[x+2];
with x=1.
Also try to solve separately the 2 recurrence equations:
RSolve[{A[k + 1] == (2k - 1) A[k] + 2k A[k - 1], A[0] == 1, A[1] == 1}, 
A[k], k]

RSolve[{B[k + 1] == (2k - 1) B[k] + 2k B[k - 1], B[0] == 0, B[1] == 1}, 
B[k], k]

and the limit of their ratio exists and it is the value of the continued 
fraction:

Limit[A[n] / B[n] , n->Infinity].

You are now busy little job ;-)

    ~Scout~






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