Re: Limit

*To*: mathgroup at smc.vnet.net*Subject*: [mg64739] Re: Limit*From*: "Scout" <Scout at nodomain.com>*Date*: Thu, 2 Mar 2006 06:47:23 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

"Scout" <Scout at nodomain.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~