Discuss the convergence of the sequence: $a_1=1,a_{n+1}=\sqrt{2+a_n} \quad
\forall n \in \mathbb{N}$ – math.stackexchange.com
Computing the first few terms $$a_1=1,
a_2=\sqrt{3}=1.732....,a_3=1.9318....,a_4=1.9828...$$ I feel that
$(a_n)_{n\in \mathbb{N}}$ is bounded above by 2, although I have no
logical reasoning for this. …
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