i am trying to calculate the limit of an:=√1+√1+√1+√1+.. with a0:=1 and an+1:=√1+an i am badly stuck not knowing how to find the limit of this sequence and where to start the proof. i did some calculations but still cannot figure out the formal way of finding the limit of this sequence. what i tried is:
(1+(1+(1+..)12)12)12 but i am totally stuck here
Answer
We (inductively) show following properties for sequence given by an+1=√1+an,a0=1
- an≥0 for all n∈N
- (an) is monotonically increasing
- (an) is bounded above by 2
Then by Monotone Convergence Theorem, the sequence converges hence the limit of sequence exists. Let lim then \lim a_{n+1} = a as well. Using Algebraic Limit Theorem, we get
\lim a_{n+1} = \sqrt{1 + \lim a_n} \implies a = \sqrt {1 + a}
Solving above equation gives out limit. Also we note that from Order Limit Theorem, we get a_n \ge 0 \implies \lim a_n \ge 0.
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