       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 - (1 + 2/(3 + 4/(5 + 6/(7 + 8))))
--> 0

Let's have a look, if it converges (and how fast):
N[cf /@ Range]
-->
{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, 150] == 1/(Sqrt[E] - 1)
--> True

If I only had paid more attention to the number theory lectures... :-\

Peter

```

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