text: # infinite sequence
  
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  • Dec 6th 2011, 11:01 PM
  jaqueh
  infinite sequence
  What does the series converge to where
  [a_1]=√12
  [a_n+1]=√(12+a_n) ?
  
  I tried:
  L = √(12+√(12+√(12+...)))
  then L^2 = 12+√(12+√(12+...))
  then L^∞ = 12^∞ + 12^∞-1...+12
  thus L = ∞√(12^∞ + 12^∞-1...+12) but i dont think that this is right.
  
  Then someone on physics forum told me that a_n can be written as √(12+a_n+1) so a_n+1 converges because it is in the series itself, but what is the limit?
  • Dec 6th 2011, 11:49 PM
  CaptainBlack
  Re: infinite series
  Quote:
  
  Originally Posted by jaqueh
  What does the series converge to where
  [a_1]=√12
  [a_n+1]=√(12+a_n) ?
  
  I tried:
  L = √(12+√(12+√(12+...)))
  then L^2 = 12+√(12+√(12+...))
  then L^∞ = 12^∞ + 12^∞-1...+12
  thus L = ∞√(12^∞ + 12^∞-1...+12) but i dont think that this is right.
  
  Then someone on physics forum told me that a_n can be written as √(12+a_n+1) so a_n+1 converges because it is in the series itself, but what is the limit?
  
  If this converges the limit should satisfy:
  
  $L=\sqrt{12+L}$
  
  squaring:
  
  $L^2=12+L$
  
  rearranging:
  
  $L^2-L-12=0$
  
  This has exactly one positive root and it is the limit you seek.
  
  CB