Consider the sequence of real number {$\displaystyle a_n$} $\displaystyle n \in N$ defined recursively by $\displaystyle a_1 = 0; a_n+1 = \sqrt{6+a_n} \forall n \geq 1$
Show that the sequence is convergent and find its limit.
Consider the sequence of real number {$\displaystyle a_n$} $\displaystyle n \in N$ defined recursively by $\displaystyle a_1 = 0; a_n+1 = \sqrt{6+a_n} \forall n \geq 1$
Show that the sequence is convergent and find its limit.
1.- prove inductively that the sequence is bounded above , by 3 say.
2.- prove inductively that a_n <= a_(n+1) (monot. ascending)
3.- thus the limit exists by a theorem. Now use a_(n+1) = sqrt(6 + a_n) and arithmetic of limits to find the limit.
wow, I was actually really close to that. I started to use induction, but then I thought I had to prove it was cauchy or something since that's all my professor forces into our head.... thanks!