Find sqrt( 2 + sqrt( 2 + sqrt( 2 + )))

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Find the limit as n tends to infinity;

sqrt(2), sqrt(2+sqrt(2)), sqrt(2+sqrt(2+sqrt(2))), etc... n times
 
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If an is the nth value, then [itex]a_{n+1}= \sqrt{2+ a_n}[/itex]

IF the sequence {an} converges (you will need to prove that, perhaps by proving that it is an increasing sequence and has an upper bound), call the limit A.
Then we must have [itex]\lim_{n\rightarrow \infty}a_{n+1}= \sqrt{2+ \lim_{n\rightarrow \infty}a_n}[/itex] so [itex]A= \sqrt{2+ A}[/itex].
 


Thank you very much.
All is clear.
 


Sorry i have a small problem

How come A=sqrt(2+A) ?
A=limit of an not a(n+1)
so Limit a(n+1)=sqrt(2+A)
No?