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~