Converging Sequence: Find the Limit of t_n

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The recursive sequence defined by t1=sqrt(2) and t_(n+1)=sqrt(2+sqrt(t_n)) is analyzed for convergence. It is suggested that as n approaches infinity, the difference between t_n and t_(n+1) becomes negligible, allowing them to be treated as equal at the limit, denoted as T. By substituting T into the recursive formula, an equation is formed that can be solved for T. The discussion emphasizes finding a non-decimal expression for the limit rather than approximating it numerically. Ultimately, the limit of the sequence can be determined by solving the derived equation for T.
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Homework Statement



Let t1=sqrt(2)

Let t_(n+1)=sqrt(2+sqrt(t_n)) (it's a recursively defined series)

What does it converge to?


Homework Equations





The Attempt at a Solution



I calculated it out for some values andI get 1.8312 (approx), but I don't want to express it in decimals, and I want to know if there's a good way to do this.
 
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Let's suppose it converges, then t_(n+1) and t_n will have a infinitesimally small difference as n approaches infinity, and you can consider them the same number. (aka the limit of this sequence)
 
I would have phrased it differently, but basically said the same thing. Since {tn+1[/b]} is exactly the same sequence as {tn}, just indexed differently, taking the limit as n goes to infinity gives the same value, say T, on both sides. Solve that equation for T. (That's not a trivial equation but there is one obvious root!)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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