August 13th, 2018, 02:08 AM  #1 
Newbie Joined: Aug 2018 From: Taiwan Posts: 1 Thanks: 0  [2;3,5,7,11,13,17,...]=?
Does anyone know how to calculate the exact value of this continued fraction [2;3,5,7,11,13,17,...] ?

August 13th, 2018, 07:03 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
My guess is that it can't be simplified.

August 22nd, 2018, 08:26 PM  #3 
Member Joined: Jul 2010 Posts: 83 Thanks: 2 
Wild guess: log(exp(1)+1)+1

August 22nd, 2018, 11:17 PM  #4 
Member Joined: Aug 2018 From: Nigeria Posts: 71 Thanks: 2 
You can using this formula S=a/1r where S=sum of terms to infity a=first term r=common ratio 
August 23rd, 2018, 06:58 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
There isn't a common ratio in this case.
