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