Re: Limit

*To*: mathgroup at smc.vnet.net*Subject*: [mg64693] Re: Limit*From*: Peter Pein <petsie at dordos.net>*Date*: Tue, 28 Feb 2006 01:49:25 -0500 (EST)*References*: <dtrvmt$lv0$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Sinan Kapçak schrieb: > 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, First write a function to approximate your continued fraction: cf[n_] := 1 + Fold[(#2 - 1)/(#1 + #2) & , 2*n + 2, Reverse[2*Range[n] + 1]] (*Test*) cf[3] - (1 + 2/(3 + 4/(5 + 6/(7 + 8)))) --> 0 Let's have a look, if it converges (and how fast): N[cf /@ Range[10]] --> {1.2857142857142858, 1.5945945945945945, 1.5346534653465347, 1.5422045680238332, 1.5414334169814963, 1.5414984960673195, 1.541493802752502, 1.5414940982569976, 1.5414940817435092, 1.5414940825731342} With the help of http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html one could guess N[1/(Sqrt[E] - 1)] -->1.5414940825367982 is the value of the c.f. At least the first 150 Digits are the same: N[cf[100], 150] == 1/(Sqrt[E] - 1) --> True If I only had paid more attention to the number theory lectures... :-\ Peter