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Re: Limit
*To*: mathgroup at smc.vnet.net
*Subject*: [mg64712] Re: Limit
*From*: Maxim <m.r at inbox.ru>
*Date*: Tue, 28 Feb 2006 05:02:15 -0500 (EST)
*References*: <dtrvmt$lv0$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On Sun, 26 Feb 2006 10:25:33 +0000 (UTC), <sinankapcak at yahoo.com> wrote:
> i want to find the value of the limit
>
> 1+(2/(3+(4/(5+(6/(7+...
>
> how can i do that with Mathematica?
>
> Tnx..
>
The answer is 1/(Sqrt[E] - 1). The derivation is given here:
http://arxiv.org/PS_cache/math/pdf/0508/0508227.pdf .
We can solve this problem in Mathematica by reformulating it in terms of
recurrence equations. The nth convergent can be constructed like this:
In[1]:= x[n_] := Fold[
With[{a = #2 + 1}, HoldForm[#2 + a/#]]&,
2*n + 1, Range[2*n - 1, 1, -2]]
In[2]:= x[3]
Out[2]= HoldForm[1 + 2/HoldForm[3 + 4/HoldForm[5 + 6/7]]]
Therefore, we can view it as the nth term in the following sequence:
f[0] = 2*n + 1;
f[k_] := (2*n - 2*k + 2)/f[k - 1] + 2*n - 2*k + 1;
and we need to find the limit as n -> Infinity. RSolve can be applied to
this equation directly, but the result will contain terms such as Gamma[n
- k], making it indeterminate for k = n. So we make a suitable change of
variables first:
In[3]:= RSolve[
{y[k] == (2*n - 2*k + 2)/y[k - 1] + 2*n - 2*k + 1, y[0] == 2*n + 1} /.
{y -> (y[n - #] - 1&), k -> n - k},
y, k]
Out[3]= {{y -> Function[{k}, ((-1)^(-k + n)*2^(1 - k + n)*Sqrt[E]*(1
+ k)*Gamma[2 + n])/(Sqrt[E]*Gamma[2 + k] + (-1)^n*2^(1 + n)*Gamma[2
+ n]*Gamma[1 + k, -(1/2)] + (-1)^n*2^(1 + n)*k*Gamma[2 + n]*Gamma[1 + k,
-(1/2)] - (-1)^n*2^(1 + n)*Gamma[2 + k]*Gamma[1 + n, -(1/2)] - (-1)^n*2^(1
+ n)*n*Gamma[2 + k]*Gamma[1 + n, -(1/2)])]}}
In[4]:= Limit[y[0] - 1 /. %[[1]], n -> Infinity]
Out[4]= 1/(-1 + Sqrt[E])
In[5]:= % - ReleaseHold //@ x[11] // N
Out[5]= -1.2234657731369225*^-13
Maxim Rytin
m.r at inbox.ru
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