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Re: Limit


"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~


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